aekaakee: Properties of Triangles

Showing posts with label Properties of Triangles. Show all posts

Properties of Triangles

In a $\triangle ABC$, prove that $\;$ $\dfrac{r_1}{bc} + \dfrac{r_2}{ca} + \dfrac{r_3}{ab} = \dfrac{1}{r} - \dfrac{1}{2 R}$

Escribed radius $= r_1 = 4 R \sin \left(\dfrac{A}{2}\right) \cos \left(\dfrac{B}{2}\right) \cos \left(\dfrac{C}{2}\right)$

Escribed radius $= r_2 = 4 R \sin \left(\dfrac{B}{2}\right) \cos \left(\dfrac{C}{2}\right) \cos \left(\dfrac{A}{2}\right)$

Escribed radius $= r_3 = 4 R \sin \left(\dfrac{C}{2}\right) \cos \left(\dfrac{A}{2}\right) \cos \left(\dfrac{B}{2}\right)$

By sine rule, $\;$ $\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} = 2 R$

$\implies$ $a = 2 R \sin A$, $\;$ $b = 2 R \sin B$, $\;$ $c = 2 R \sin C$

where $\;$ $R$ $\;$ is the circumradius of $\triangle ABC$

Now,

$\begin{aligned} \dfrac{r_1}{bc} & = \dfrac{4 R \sin \left(\dfrac{A}{2}\right) \cos \left(\dfrac{B}{2}\right) \cos \left(\dfrac{C}{2}\right)}{2 R \sin B \times 2 R \sin C} \\\\ & \left[\text{Note: } \sin \theta = 2 \sin \left(\dfrac{\theta}{2}\right) \cos \left(\dfrac{\theta}{2}\right) \right] \\\\ & = \dfrac{\sin \left(\dfrac{A}{2}\right) \cos \left(\dfrac{B}{2}\right) \cos \left(\dfrac{C}{2}\right)}{R \times 2 \sin \left(\dfrac{B}{2}\right) \cos \left(\dfrac{B}{2}\right) \times 2 \sin \left(\dfrac{C}{2}\right) \cos \left(\dfrac{C}{2}\right)} \\\\ & = \dfrac{\sin \left(\dfrac{A}{2}\right)}{4 R \sin \left(\dfrac{B}{2}\right) \sin \left(\dfrac{C}{2}\right)} \\\\ & = \dfrac{\sin^2 \left(\dfrac{A}{2}\right)}{4 R \sin \left(\dfrac{A}{2}\right) \sin \left(\dfrac{B}{2}\right) \sin \left(\dfrac{C}{2}\right)} \\\\ & = \dfrac{\sin^2 \left(\dfrac{A}{2}\right)}{r} \;\;\; \cdots \; (1a) \end{aligned}$

Similarly,

$\dfrac{r_2}{ca} = \dfrac{\sin^2 \left(\dfrac{B}{2}\right)}{r}$ $\;\;\; \cdots \; (1b)$ $\;$ and $\;$ $\dfrac{r_3}{ab} = \dfrac{\sin^2 \left(\dfrac{C}{2}\right)}{r}$ $\;\;\; \cdots \; (1c)$

$\therefore \;$ In view of equations $(1a)$, $(1b)$ and $(1c)$, we have

$\begin{aligned} LHS & = \dfrac{r_1}{bc} + \dfrac{r_2}{ca} + \dfrac{r_3}{ab} \\\\ & = \dfrac{1}{r} \left[\sin^2 \left(\dfrac{A}{2}\right) + \sin^2 \left(\dfrac{B}{2}\right) + \sin^2 \left(\dfrac{C}{2}\right)\right] \;\;\; \cdots \; (2) \end{aligned}$

$\left[\text{Note: } \sin^2 \left(\dfrac{\theta}{2}\right) = \dfrac{1 - \cos \theta}{2}\right]$

Now,

$\sin^2 \left(\dfrac{A}{2}\right) + \sin^2 \left(\dfrac{B}{2}\right) + \sin^2 \left(\dfrac{C}{2}\right)$

$= \dfrac{1 - \cos A}{2} + \dfrac{1 - \cos B}{2} + \sin^2 \left(\dfrac{C}{2}\right)$

$= 1 - \dfrac{1}{2} \left[\cos A + \cos B\right] + \sin^2 \left(\dfrac{C}{2}\right)$

$\left[\text{Note: }\cos \alpha + \cos \beta = 2 \cos \left(\dfrac{\alpha + \beta}{2}\right) \cos \left(\dfrac{\alpha - \beta}{2}\right)\right]$

$= 1 - \cos \left(\dfrac{A + B}{2}\right) \cos \left(\dfrac{A - B}{2}\right) + \sin^2 \left(\dfrac{C}{2}\right)$

$\left[\text{Note: In } \triangle ABC, \; A + B + C = \pi \implies \dfrac{A + B}{2} = \dfrac{\pi}{2} - \dfrac{C}{2} \right.$
$\left. \therefore \; \cos \left(\dfrac{A + B}{2}\right) = \cos \left(\dfrac{\pi}{2} - \dfrac{C}{2}\right) = \sin \left(\dfrac{C}{2}\right) \right]$

$\therefore \; \sin^2 \left(\dfrac{A}{2}\right) + \sin^2 \left(\dfrac{B}{2}\right) + \sin^2 \left(\dfrac{C}{2}\right)$

$= 1 - \sin \left(\dfrac{C}{2}\right) \cos \left(\dfrac{A - B}{2}\right) + \sin^2 \left(\dfrac{C}{2}\right)$

$= 1 - \sin \left(\dfrac{C}{2}\right) \left[\cos \left(\dfrac{A - B}{2}\right) - \sin \left(\dfrac{C}{2}\right)\right]$

$\left[\text{Note: In } \triangle ABC, \; A + B + C = \pi \implies \dfrac{C}{2} = \dfrac{\pi}{2} - \left(\dfrac{A + B}{2}\right) \right.$
$\left. \therefore \; \sin \left(\dfrac{C}{2}\right) = \sin \left[\dfrac{\pi}{2} - \left(\dfrac{A + B}{2}\right)\right] = \cos \left(\dfrac{A + B}{2}\right) \right]$

$\therefore \; \sin^2 \left(\dfrac{A}{2}\right) + \sin^2 \left(\dfrac{B}{2}\right) + \sin^2 \left(\dfrac{C}{2}\right)$

$= 1 - \sin \left(\dfrac{C}{2}\right) \left[\cos \left(\dfrac{A - B}{2}\right) - \cos \left(\dfrac{A + B}{2}\right)\right]$

$\left[\text{Note: }\cos \alpha - \cos \beta = 2 \sin \left(\dfrac{\alpha + \beta}{2}\right) \sin \left(\dfrac{\beta - \alpha}{2}\right)\right]$

$\therefore \; \sin^2 \left(\dfrac{A}{2}\right) + \sin^2 \left(\dfrac{B}{2}\right) + \sin^2 \left(\dfrac{C}{2}\right)$

$= 1 - \sin \left(\dfrac{C}{2}\right) \times 2 \sin \left(\dfrac{A - B + A + B}{4}\right) \sin \left(\dfrac{A - B - A + B}{4}\right)$

$= 1 - 2 \sin \left(\dfrac{A}{2}\right) \sin \left(\dfrac{B}{2}\right) \sin \left(\dfrac{C}{2}\right)$ $\;\;\; \cdots \; (3)$

$\therefore \;$ We have from equations $(2)$ and $(3)$,

$\begin{aligned} LHS & = \dfrac{1}{r} \left[1 - 2 \sin \left(\dfrac{A}{2}\right) \sin \left(\dfrac{B}{2}\right) \sin \left(\dfrac{C}{2}\right)\right] \\\\ & \left[\text{Note: In-radius } = r = 4 R \sin \left(\dfrac{A}{2}\right) \sin \left(\dfrac{B}{2}\right) \sin \left(\dfrac{C}{2}\right) \right. \\ & \left. \implies \dfrac{r}{2 R} = 2 \sin \left(\dfrac{A}{2}\right) \sin \left(\dfrac{B}{2}\right) \sin \left(\dfrac{C}{2}\right) \right] \\\\ & = \dfrac{1}{r} \left[1 - \dfrac{r}{2 R}\right] \\\\ & = \dfrac{1}{r} - \dfrac{1}{2 R} = RHS \end{aligned}$

Hence proved.

Properties of Triangles

In a $\triangle ABC$, prove that $\;$ $\left(r_1 + r_2\right) \tan \left(\dfrac{C}{2}\right) = \left(r_3 - r\right) \cot \left(\dfrac{C}{2}\right) = c$

Escribed radius $= r_1 = \dfrac{\Delta}{s - a}$

Escribed radius $= r_2 = \dfrac{\Delta}{s - b}$

$\tan \left(\dfrac{C}{2}\right) = \dfrac{\Delta}{s \left(s - c\right)}$

$\Delta = \sqrt{s \left(s - a\right) \left(s - b\right) \left(s - c\right)}$ $\;$ is the area of $\triangle ABC$

$\implies$ $\Delta^2 = s \left(s - a\right) \left(s - b\right) \left(s - c\right)$

$s = \dfrac{a + b + c}{2}$ $\;$ is the semi-perimeter of $\triangle ABC$

$\implies$ $2 s = a + b + c$

$\begin{aligned} \therefore \; \left(r_1 + r_2\right) \tan \left(\dfrac{C}{2}\right) & = \left[\dfrac{\Delta}{s - a} + \dfrac{\Delta}{s - b}\right] \times \dfrac{\Delta}{s \left(s - c\right)} \\\\ & = \Delta^2 \times \left[\dfrac{s - b + s - a}{s \left(s - a\right) \left(s - b\right) \left(s - c\right)}\right] \\\\ & = 2 s - a - b \\\\ & = a + b + c - a - b = c \;\;\; \cdots \; (1) \end{aligned}$

Escribed radius $= r_3 = s \tan \left(\dfrac{C}{2}\right)$

In-radius $= r = \left(s - c\right) \tan \left(\dfrac{C}{2}\right)$

$\begin{aligned} \therefore \; \left(r_3 - r\right) \cot \left(\dfrac{C}{2}\right) & = r_3 \cot \left(\dfrac{C}{2}\right) - r \cot \left(\dfrac{C}{2}\right) \\\\ & = s \tan \left(\dfrac{C}{2}\right) \times \cot \left(\dfrac{C}{2}\right) - \left(s - c\right) \tan \left(\dfrac{C}{2}\right) \times \cot \left(\dfrac{C}{2}\right) \\\\ & = s - \left(s - c\right) = c \;\;\; \cdots \; (2) \end{aligned}$

$\therefore \;$ We have from equations $(1)$ and $(2)$

$\left(r_1 + r_2\right) \tan \left(\dfrac{C}{2}\right) = \left(r_3 - r\right) \cot \left(\dfrac{C}{2}\right) = c$

Hence proved.

Properties of Triangles

In a $\triangle ABC$, prove that $\;$ $r r_1 \cot \left(\dfrac{A}{2}\right) = \Delta$

Circumradius $= R = \dfrac{a b c}{4 \Delta}$ $\;$ where $\Delta$ is the area of $\triangle ABC$

i.e. $\;$ $\Delta = \dfrac{a b c}{4 R}$

Inradius $= r = \dfrac{a \sin \left(\dfrac{B}{2}\right) \sin \left(\dfrac{C}{2}\right)}{\cos \left(\dfrac{A}{2}\right)}$

Escribed radius $= r_1 = \dfrac{a \cos \left(\dfrac{B}{2}\right) \cos \left(\dfrac{C}{2}\right)}{\cos \left(\dfrac{A}{2}\right)}$

$\begin{aligned} LHS & = r r_1 \cot \left(\dfrac{A}{2}\right) \\\\ & = \dfrac{a \sin \left(\dfrac{B}{2}\right) \sin \left(\dfrac{C}{2}\right)}{\cos \left(\dfrac{A}{2}\right)} \times \dfrac{a \cos \left(\dfrac{B}{2}\right) \cos \left(\dfrac{C}{2}\right)}{\cos \left(\dfrac{A}{2}\right)} \times \cot \left(\dfrac{A}{2}\right) \\\\ & = \dfrac{a^2 \sin \left(\dfrac{B}{2}\right) \cos \left(\dfrac{B}{2}\right) \sin \left(\dfrac{C}{2}\right) \cos \left(\dfrac{C}{2}\right)}{\cos^2 \left(\dfrac{A}{2}\right)} \times \dfrac{\cos \left(\dfrac{A}{2}\right)}{\sin \left(\dfrac{A}{2}\right)} \\\\ & \left[\text{Note: } \dfrac{\sin \theta}{2} = \sin \left(\dfrac{\theta}{2}\right) \cos \left(\dfrac{\theta}{2}\right)\right] \\\\ & = a^2 \times \dfrac{\sin B}{2} \times \dfrac{\sin C}{2} \times \dfrac{2}{\sin A} \\\\ & \left[\text{Note: By sine rule, } \dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} = 2R \right. \\ & \left. \implies \sin A = \dfrac{a}{2 R}, \; \sin B = \dfrac{b}{2 R}, \; \sin C = \dfrac{c}{2 R} \right] \\\\ & = a^2 \times \dfrac{1}{2} \times \dfrac{b}{2 R} \times \dfrac{c}{2 R} \times \dfrac{2 R}{a} \\\\ & = \dfrac{a b c}{4 R} \\\\ & = \Delta = RHS \end{aligned}$

Hence proved.

Properties of Triangles

In a $\triangle ABC$, prove that $\;$ $\dfrac{1}{r^2} + \dfrac{1}{r_1^2} + \dfrac{1}{r_2^2} + \dfrac{1}{r_3^2} = \dfrac{a^2 + b^2 + c^2}{\Delta^2}$

Properties of Triangles

In a $\triangle ABC$, prove that $\;$ $a \left(r r_1 + r_2 r_3\right) = b \left(r r_2 + r_3 r_1\right) = c \left(r r_3 + r_1 r_2\right)$

Inradius $= r = \dfrac{\Delta}{s}$

Escribed radius $= r_1 = \dfrac{\Delta}{s - a}$

Escribed radius $= r_2 = \dfrac{\Delta}{s - b}$

Escribed radius $= r_3 = \dfrac{\Delta}{s - c}$

where $\;$ $\Delta = \sqrt{s \left(s - a\right) \left(s - b\right) \left(s - c\right)}$ $\;$ is the area of $\triangle ABC$

i.e. $\;$ $\Delta^2 = s \left(s - a\right) \left(s - b\right) \left(s - c\right)$

and $\;$ $s = \dfrac{a + b + c}{2}$ $\;$ is the semi-perimeter of $\triangle ABC$

i.e. $\;$ $2 s = a + b + c$

Now,

$\begin{aligned} a \left(r r_1 + r_2 r_3\right) & = a \left[\dfrac{\Delta}{s} \times \dfrac{\Delta}{\left(s - a\right)} + \dfrac{\Delta}{\left(s - b\right)} \times \dfrac{\Delta}{\left(s - c\right)}\right] \\\\ & = \dfrac{a \Delta^2 \left[s^2 - sc - s b + b c + s^2 - s a\right]}{s \left(s - a\right) \left(s - b\right) \left(s - c \right)} \\\\ & = a \left[2 s^2 - s \left(a + b + c\right) + bc\right] \\\\ & = a \left[2 s^2 - 2 s^2 + bc\right] \\\\ & = a b c \;\;\; \cdots \; (1a) \end{aligned}$

$\begin{aligned} b \left(r r_2 + r_3 r_1\right) & = b \left[\dfrac{\Delta}{s} \times \dfrac{\Delta}{\left(s - b\right)} + \dfrac{\Delta}{\left(s - c\right)} \times \dfrac{\Delta}{\left(s - a\right)}\right] \\\\ & = \dfrac{b \Delta^2 \left[s^2 - s a - s c + a c + s^2 - s b\right]}{s \left(s - a\right) \left(s - b\right) \left(s - c \right)} \\\\ & = b \left[2 s^2 - s \left(a + b + c\right) + ac\right] \\\\ & = b \left[2 s^2 - 2 s^2 + ac\right] \\\\ & = a b c \;\;\; \cdots \; (1b) \end{aligned}$

$\begin{aligned} c \left(r r_3 + r_1 r_2\right) & = c \left[\dfrac{\Delta}{s} \times \dfrac{\Delta}{\left(s - c\right)} + \dfrac{\Delta}{\left(s - a\right)} \times \dfrac{\Delta}{\left(s - b\right)}\right] \\\\ & = \dfrac{c \Delta^2 \left[s^2 - s b - s a + a b + s^2 - s c\right]}{s \left(s - a\right) \left(s - b\right) \left(s - c \right)} \\\\ & = c \left[2 s^2 - s \left(a + b + c\right) + ab\right] \\\\ & = c \left[2 s^2 - 2 s^2 + ab\right] \\\\ & = a b c \;\;\; \cdots \; (1c) \end{aligned}$

$\therefore \;$ We have from equations $(1a)$, $(1b)$ and $(1c)$,

$a \left(r r_1 + r_2 r_3\right) = b \left(r r_2 + r_3 r_1\right) = c \left(r r_3 + r_1 r_2\right)$

Hence proved.

Properties of Triangles

In a $\triangle ABC$, prove that $\;$ $\dfrac{r \; r_1}{r_2 \; r_3} = \tan^2 \left(\dfrac{A}{2}\right)$

Inradius $= r = 4 R \sin \left(\dfrac{A}{2}\right) \sin \left(\dfrac{B}{2}\right) \sin \left(\dfrac{C}{2}\right)$

Escribed radius $= r_1 = 4 R \sin \left(\dfrac{A}{2}\right) \cos \left(\dfrac{B}{2}\right) \cos \left(\dfrac{C}{2}\right)$

Escribed radius $= r_2 = 4 R \sin \left(\dfrac{B}{2}\right) \cos \left(\dfrac{C}{2}\right) \cos \left(\dfrac{A}{2}\right)$

Escribed radius $= r_3 = 4 R \sin \left(\dfrac{C}{2}\right) \cos \left(\dfrac{A}{2}\right) \cos \left(\dfrac{B}{2}\right)$

where $\;$ $R$ $\;$ is the circumradius of $\triangle ABC$.

Now,

$\begin{aligned} rr_1 & = 4 R \sin \left(\dfrac{A}{2}\right) \sin \left(\dfrac{B}{2}\right) \sin \left(\dfrac{C}{2}\right) \times 4 R \sin \left(\dfrac{A}{2}\right) \cos \left(\dfrac{B}{2}\right) \cos \left(\dfrac{C}{2}\right) \\\\ & = 4 R^2 \sin^2 \left(\dfrac{A}{2}\right) \times 2 \sin \left(\dfrac{B}{2}\right) \cos \left(\dfrac{B}{2}\right) \times 2 \sin \left(\dfrac{C}{2}\right) \cos \left(\dfrac{C}{2}\right) \\\\ & \left[\text{Note: }\sin \theta = 2 \sin \left(\dfrac{\theta}{2}\right) \cos \left(\dfrac{\theta}{2}\right)\right] \\\\ & = 4 R^2 \sin^2 \left(\dfrac{A}{2}\right) \sin B \sin C \;\;\; \cdots \; (1a) \end{aligned}$

$\begin{aligned} r_2 r_3 & = 4 R \sin \left(\dfrac{B}{2}\right) \cos \left(\dfrac{C}{2}\right) \cos \left(\dfrac{A}{2}\right) \times 4 R \sin \left(\dfrac{C}{2}\right) \cos \left(\dfrac{A}{2}\right) \cos \left(\dfrac{B}{2}\right) \\\\ & = 4 R^2 \cos^2 \left(\dfrac{A}{2}\right) \times 2 \sin \left(\dfrac{B}{2}\right) \cos \left(\dfrac{B}{2}\right) \times 2 \sin \left(\dfrac{C}{2}\right) \cos \left(\dfrac{C}{2}\right) \\\\ & = 4 R^2 \cos^2 \left(\dfrac{A}{2}\right) \sin B \sin C \;\;\; \cdots \; (1b) \end{aligned}$

$\therefore \;$ We have from equations $(1a)$ and $(1b)$,

$\begin{aligned} LHS = \dfrac{r r_1}{r_2 r_3} & = \dfrac{4 R^2 \sin^2 \left(\dfrac{A}{2}\right) \sin B \sin C}{4 R^2 \cos^2 \left(\dfrac{A}{2}\right) \sin B \sin C} \\\\ & = \dfrac{\sin^2 \left(\dfrac{A}{2}\right)}{\cos^2 \left(\dfrac{A}{2}\right)} \\\\ & = \tan^2 \left(\dfrac{A}{2}\right) = RHS \end{aligned}$

Hence proved.

Properties of Triangles

In a $\triangle ABC$, prove that $\;$ $\Delta = 4 \; R \; r \; \cos \left(\dfrac{A}{2}\right) \; \cos \left(\dfrac{B}{2}\right) \; \cos \left(\dfrac{C}{2}\right)$

In-radius $\;$ $r = 4 R \sin \left(\dfrac{A}{2}\right) \sin \left(\dfrac{B}{2}\right) \sin \left(\dfrac{C}{2}\right)$

Circumradius $\;$ $R = \dfrac{a \; b \; c}{4 \Delta}$

$\sin A = \dfrac{2 \Delta}{b c}$, $\;$ $\sin B = \dfrac{2 \Delta}{c a}$, $\;$ $\sin C = \dfrac{2 \Delta}{a b}$

where $\;$ $\Delta$ $\;$ is the area of $\triangle ABC$.

$\begin{aligned} RHS & = 4 \; R \; r \; \cos \left(\dfrac{A}{2}\right) \; \cos \left(\dfrac{B}{2}\right) \; \cos \left(\dfrac{C}{2}\right) \\\\ & = 4 R \times 4 R \sin \left(\dfrac{A}{2}\right) \sin \left(\dfrac{B}{2}\right) \sin \left(\dfrac{C}{2}\right) \times \cos \left(\dfrac{A}{2}\right) \cos \left(\dfrac{B}{2}\right) \cos \left(\dfrac{C}{2}\right) \\\\ & \left[\text{Note: } \sin \left(\dfrac{\theta}{2}\right) \cos \left(\dfrac{\theta}{2}\right) = \dfrac{\sin \theta}{2}\right] \\\\ & = 16 R^2 \times \dfrac{\sin A}{2} \times \dfrac{\sin B}{2} \times \dfrac{\sin C}{2} \\\\ & = 2 R^2 \sin A \sin B \sin C \\\\ & = 2 \times \left(\dfrac{abc}{4 \Delta}\right)^2 \times \left(\dfrac{2 \Delta}{bc}\right) \times \left(\dfrac{2 \Delta}{ca}\right) \times \left(\dfrac{2 \Delta}{ab}\right) \\\\ & = \dfrac{16 \; \Delta^3 \; a^2 \; b^2 \; c^2}{16 \; \Delta^2 \; a^2 \; b^2 \; c^2} \\\\ & = \Delta = LHS \end{aligned}$

Hence proved.

Properties of Triangles

In a $\triangle ABC$, prove that $\;$ $\dfrac{1}{r_1} + \dfrac{1}{r_2} + \dfrac{1}{r_3} - \dfrac{1}{r} = 0$

Escribed radius $\;$ $r_1 = \dfrac{\Delta}{s - a}$

Escribed radius $\;$ $r_2 = \dfrac{\Delta}{s - b}$

Escribed radius $\;$ $r_3 = \dfrac{\Delta}{s - c}$

In-radius $\;$ $r = \dfrac{\Delta}{s}$

where $\;$ $\Delta$ $\;$ is the area of $\triangle ABC$

and $\;$ $s = \dfrac{a + b + c}{2}$ $\;$ is the semi-perimeter of $\triangle ABC$

$\implies$ $2 s = a + b + c$

$\begin{aligned} LHS & = \dfrac{1}{r_1} + \dfrac{1}{r_2} + \dfrac{1}{r_3} - \dfrac{1}{r} \\\\ & = \dfrac{s - a}{\Delta} + \dfrac{s - b}{\Delta} + \dfrac{s - c}{\Delta} - \dfrac{s}{\Delta} \\\\ & = \dfrac{1}{\Delta} \left[s - a + s - b + s - c - s\right] \\\\ & = \dfrac{1}{\Delta} \left[2 s - \left(a + b + c\right)\right] \\\\ & = \dfrac{1}{\Delta} \left[\left(a + b + c\right) - \left(a + b + c\right)\right] \\\\ & = 0 = RHS \end{aligned}$

Hence proved.

Properties of Triangles

In a $\triangle ABC$, prove that $\;$ $r_1 \cdot r_2 \cdot r_3 = r^3 \cot^2 \left(\dfrac{A}{2}\right) \cot^2 \left(\dfrac{B}{2}\right) \cot^2 \left(\dfrac{C}{2}\right)$ $\;$ where the symbols have their usual meanings.

Properties of Triangles

In a $\triangle ABC$, prove that $\;$ $r_1 + r_2 + r_3 - r = 4R$ $\;$ where the symbols have their usual meanings.

Escribed radius $\;$ $r_1 = 4 R \sin \left(\dfrac{A}{2}\right) \cos \left(\dfrac{B}{2}\right) \cos \left(\dfrac{C}{2}\right)$

Escribed radius $\;$ $r_2 = 4 R \sin \left(\dfrac{B}{2}\right) \cos \left(\dfrac{C}{2}\right) \cos \left(\dfrac{A}{2}\right)$

Escribed radius $\;$ $r_3 = 4 R \sin \left(\dfrac{C}{2}\right) \cos \left(\dfrac{A}{2}\right) \cos \left(\dfrac{B}{2}\right)$

Inradius $\;$ $r = 4 R \sin \left(\dfrac{A}{2}\right) \sin \left(\dfrac{B}{2}\right) \sin \left(\dfrac{C}{2}\right)$

where $\;$ $R$ is the circumradius of $\triangle ABC$.

$\begin{aligned} LHS & = r_1 + r_2 + r_3 - r \\\\ & = \left(r_1 - r\right) + \left(r_2 + r_3\right) \\\\ & = 4 R \sin \left(\dfrac{A}{2}\right) \left[\cos \left(\dfrac{B}{2}\right) \cos \left(\dfrac{C}{2}\right) - \sin \left(\dfrac{B}{2}\right) \sin \left(\dfrac{C}{2}\right)\right] \\ & \hspace{1cm} + 4 R \cos \left(\dfrac{A}{2}\right) \left[\sin \left(\dfrac{B}{2}\right) \cos \left(\dfrac{C}{2}\right) + \cos \left(\dfrac{B}{2}\right) \sin \left(\dfrac{C}{2}\right)\right] \;\;\; \cdots \; (1) \end{aligned}$

Now, $\;$ $\cos \left(\alpha + \beta\right) = \cos \alpha \cos \beta - \sin \alpha \sin \beta$ $\;\;\; \cdots \; (2a)$

$\sin \left(\alpha + \beta\right) = \sin \alpha \cos \beta + \cos \alpha \sin \beta$ $\;\;\; \cdots \; (2b)$

In view of equations $(2a)$ and $(2b)$ equation $(1)$ becomes,

$LHS = 4 R \sin \left(\dfrac{A}{2}\right) \cos \left(\dfrac{B + C}{2}\right) + 4 R \cos \left(\dfrac{A}{2}\right) \sin \left(\dfrac{B + C}{2}\right)$ $\;\;\; \cdots \; (3)$

In $\triangle ABC$, $\;$ $A + B + C = \pi$

$\implies$ $B + C = \pi - A$

i.e. $\;$ $\dfrac{B + C}{2} = \dfrac{\pi}{2} - \dfrac{A}{2}$

$\therefore \;$ $\cos \left(\dfrac{B + C}{2}\right) = \cos \left(\dfrac{\pi}{2} - \dfrac{A}{2}\right) = \sin \left(\dfrac{A}{2}\right)$ $\;\;\; \cdots \; (4a)$

Similarly, $\;$ $\sin \left(\dfrac{B + C}{2}\right) = \sin \left(\dfrac{\pi}{2} - \dfrac{A}{2}\right) = \cos \left(\dfrac{A}{2}\right)$ $\;\;\; \cdots \; (4b)$

In view of equations $(4a)$ and $(4b)$ equation $(3)$ becomes,

$\begin{aligned} LHS & = 4 R \sin \left(\dfrac{A}{2}\right) \times \sin \left(\dfrac{A}{2}\right) + 4 R \cos \left(\dfrac{A}{2}\right) \times \cos \left(\dfrac{A}{2}\right) \\\\ & = 4 R \left[\sin^2 \left(\dfrac{A}{2}\right) + \cos^2 \left(\dfrac{A}{2}\right)\right] \\\\ & = 4 R = RHS \end{aligned}$

Hence proved.

Properties of Triangles

In a $\triangle ABC$, if $a = 13$, $b = 4$ and $\cos C = - \dfrac{5}{13}$, find $R$, $r$, $r_1$, $r_2$ and $r_3$.

Given: $\;$ In $\triangle ABC$, $\;$ $a = 13$, $\;$ $b = 4$, $\;$ $\cos C = - \dfrac{5}{13}$

By cosine rule, $\;$ $\cos C = \dfrac{a^2 + b^2 - c^2}{2 a b}$

$\implies$ $c^2 = a^2 + b^2 - 2 a b \cos C$

i.e. $\;$ $c^2 = \left(13\right)^2 + \left(4\right)^2 - 2 \times 13 \times 4 \times \left(- \dfrac{5}{13}\right)$

i.e. $\;$ $c^2 = 169 + 16 + 40 = 225$ $\implies$ $c = 15$

Also, $\;$ $\sin C = \sqrt{1 - \cos^2 C}$

i.e. $\;$ $\sin C = \sqrt{1 - \left(- \dfrac{5}{13}\right)^2} = \sqrt{\dfrac{144}{169}} = \dfrac{12}{13}$

Now, circumradius $\;$ $R = \dfrac{c}{\sin C} = \dfrac{15}{2 \times \dfrac{12}{13}} = 8.125$

Semi-perimeter of $\triangle ABC$ $\;$ $= s = \dfrac{a + b + c}{2} = \dfrac{13 + 4 + 15}{2} = 16$

Area of $\triangle ABC$ $\;$ $= \Delta = \dfrac{1}{2}a b \sin C = \dfrac{1}{2} \times 13 \times 4 \times \dfrac{12}{13} = 24$

Now, inradius $\;$ $r = \dfrac{\Delta}{s} = \dfrac{24}{16} = 1.5$

Escribed radius $\;$ $r_1 = \dfrac{\Delta}{s - a} = \dfrac{24}{16 - 13} = 8$

Escribed radius $\;$ $r_2 = \dfrac{\Delta}{s - b} = \dfrac{24}{16 - 4} = 2$

Escribed radius $\;$ $r_3 = \dfrac{\Delta}{s - c} = \dfrac{24}{16 - 15} = 24$

Properties of Triangles

In a triangle whose sides are $18$, $24$ and $30 \; cm$ respectively, find the circumradius, the inradius and the radii of the three escribed circles.

Let $ABC$ be the given triangle with sides $\;$ $a = 18 \; cm$, $\;$ $b = 24 \; cm$, $\;$ $c = 30 \; cm$

Semi-perimeter of $\triangle ABC$ $= s = \dfrac{a + b + c}{2} = \dfrac{18 + 24 + 30}{2} = 36 \; cm$

Area of $\triangle ABC$ $= \sqrt{s \left(s - a\right) \left(s - b\right) \left(s - c\right)}$

i.e. $\;$ $\Delta = \sqrt{36 \left(36 - 18\right) \left(36 - 24\right) \left(36 - 30\right)}$

i.e. $\;$ $\Delta = \sqrt{36 \times 18 \times 12 \times 6} = 216 \; cm^2$

Circumradius $R = \dfrac{abc}{4 \Delta} = \dfrac{18 \times 24 \times 30}{4 \times 216} = 15 \; cm$

Inradius $r = \dfrac{\Delta}{s} = \dfrac{216}{36} = 6 \; cm$

Escribed radius $r_1 = \dfrac{\Delta}{s - a} = \dfrac{216}{36 - 18} = 12 \; cm$

Escribed radius $r_2 = \dfrac{\Delta}{s - b} = \dfrac{216}{36 - 24} = 18 \; cm$

Escribed radius $r_3 = \dfrac{\Delta}{s - c} = \dfrac{216}{36 - 30} = 36 \; cm$

Properties of Triangles

The lengths of two sides of a triangle are $1$ and $\sqrt{2}$ units respectively and the angle opposite the shorter side is $30^\circ$. Prove that there are two triangles satisfying these conditions, find their angles and show that their areas are in the ratio $\sqrt{3} + 1 : \sqrt{3} - 1$

Let the sides of the triangle be $\;$ $a = 1$ unit, $\;$ $b = \sqrt{2}$ units

Angle opposite shorter side is $\;$ $A = 30^\circ$

Now, $\;$ $b \sin A = \sqrt{2} \times \sin \left(30^\circ\right) = \dfrac{\sqrt{2}}{2} = \dfrac{1}{\sqrt{2}} < a$

$\implies$ There are two values of $B$, $\;$ $0^\circ < B < 90^\circ$ (first quadrant) and $90^\circ < B < 180^\circ$ (second quadrant)

By sine rule in $\triangle ABC$, $\;$ $\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}$

$\implies$ $\sin B = \dfrac{b \sin A}{a} = \dfrac{\sqrt{2} \times \sin \left(30^\circ\right)}{1} = \dfrac{\sqrt{2}}{2} = \dfrac{1}{\sqrt{2}}$

$\implies$ $B = \sin^{-1} \left(\dfrac{1}{\sqrt{2}}\right)$

i.e. $\;$ $B = 45^\circ$ $\;$ or $\;$ $B = 180^\circ - 45^\circ = 135^\circ$

Now, in $\triangle ABC$, $A + B + C = 180^\circ$ $\implies$ $C = 180^\circ - \left(A + B\right)$

When $\;$ $B = 45^\circ$, $\;$ $C = 180^\circ - \left(30^\circ + 45^\circ\right) = 105^\circ$

When $\;$ $B = 135^\circ$, $\;$ $C = 180^\circ - \left(30^\circ + 135^\circ\right) = 15^\circ$

When $\;$ $a = 1$, $\;$ $b = \sqrt{2}$, $\;$ $A = 30^\circ$, $\;$ $B_1 = 45^\circ$, $\;$ $C_1 = 105^\circ$

Area of $\triangle A B_1 C_1$ $= \Delta_1 = \dfrac{1}{2} a \; b \sin C_1$

i.e. $\;$ $\Delta_1 = \dfrac{1}{2} \times 1 \times \sqrt{2} \times \sin \left(105^\circ\right)$

i.e. $\;$ $\Delta_1 = \dfrac{\sqrt{2}}{2} \times \dfrac{\left(\sqrt{3} + 1\right)}{2 \sqrt{2}}$

i.e. $\;$ $\Delta_1 = \dfrac{\sqrt{3} + 1}{4}$

When $\;$ $a = 1$, $\;$ $b = \sqrt{2}$, $\;$ $A = 30^\circ$, $\;$ $B_2 = 135^\circ$, $\;$ $C_2 = 15^\circ$

Area of $\triangle A B_2 C_2$ $= \Delta_2 = \dfrac{1}{2} a \; b \sin C_2$

i.e. $\;$ $\Delta_2 = \dfrac{1}{2} \times 1 \times \sqrt{2} \times \sin \left(15^\circ\right)$

i.e. $\;$ $\Delta_2 = \dfrac{\sqrt{2}}{2} \times \dfrac{\left(\sqrt{3} - 1\right)}{2 \sqrt{2}}$

i.e. $\;$ $\Delta_2 = \dfrac{\sqrt{3} - 1}{4}$

$\therefore \;$ $\Delta_1 : \Delta_2 = \dfrac{\sqrt{3} + 1}{4} : \dfrac{\sqrt{3} - 1}{4}$

i.e. $\;$ $\Delta_1 : \Delta_2 = \sqrt{3} + 1 : \sqrt{3} - 1$

Properties of Triangles

The sides of a triangle are in A.P and its area is $\dfrac{3}{5}$ths of an equilateral triangle of the same perimeter. Prove that its sides are in the ratio $3 : 5 : 7$ and find the greatest angle of the triangle.

Let the sides of the triangle be $s_1$, $s_2$ and $s_3$.

Given: $\;$ $s_1$, $s_2$ and $s_3$ are in A.P

$\therefore \;$ Let $s_1 = a - x$, $s_2 = a$, $s_3 = a + x$

Semi-perimeter of the triangle $= s = \dfrac{a - x + a + a + x}{2} = \dfrac{3a}{2}$

Area of the triangle $= \Delta = \sqrt{s \left(s - s_1\right) \left(s - s_2\right) \left(s - s_3\right)}$

i.e. $\;$ $\Delta = \sqrt{\left(\dfrac{3a}{2}\right) \left(\dfrac{3a}{2} - a + x\right) \left(\dfrac{3a}{2} - a\right) \left(\dfrac{3a}{2} - a - x\right)}$

i.e. $\;$ $\Delta = \sqrt{\left(\dfrac{3a}{2}\right) \left(\dfrac{a + 2x}{2}\right) \left(\dfrac{a}{2}\right) \left(\dfrac{a - 2x}{2}\right)}$

i.e. $\;$ $\Delta = \dfrac{a}{4} \sqrt{3 \left(a^2 - 4 x^2\right)}$ $\;\;\; \cdots \; (1)$

Area of an equilateral triangle with perimeter same as that of triangle with sides $s_1$, $s_2$ and $s_3$ is

$\Delta_1 = \dfrac{a^2 \sqrt{3}}{4}$ $\;\;\; \cdots \; (2)$

Given: $\;$ $\Delta = \dfrac{3}{5} \Delta_1$

$\therefore \;$ In view of equations $(1)$ and $(2)$ we have,

$\dfrac{a}{4} \sqrt{3 \left(a^2 - 4 x^2\right)} = \dfrac{3}{5} \times \dfrac{a^2 \sqrt{3}}{4}$

i.e. $\;$ $\sqrt{a^2 - 4x^2} = \dfrac{3 a}{5}$

i.e. $\;$ $a^2 - 4 x^2 = \dfrac{9 a^2}{25}$

i.e. $\;$ $4 x^2 = a^2 - \dfrac{9a^2}{25} = \dfrac{16 a^2}{25}$

i.e. $\;$ $x^2 = \dfrac{4 a^2}{25}$

$\implies$ $x = \dfrac{2 a}{5}$ $\;\;\;$ [negative value for $x$ is rejected as sides of a triangle cannot be negative]

$\therefore \;$ The sides of the triangle are

$s_1 = a - \dfrac{2a}{5}$, $\;$ $s_2 = a$, $\;$ $s_3 = a + \dfrac{2a}{5}$

i.e. $\;$ $s_1 = \dfrac{3a}{5}$, $\;$ $s_2 = a$, $\;$ $s_3 = \dfrac{7a}{5}$

Now, ratio of sides of the triangle is

$s_1 : s_2 : s_3 = \dfrac{3a}{5} : a : \dfrac{7a}{5}$

i.e. $\;$ $s_1 : s_2 : s_3 = 3 : 5 : 7$

Greatest side of the triangle is $s_3 = \dfrac{7a}{5}$

$\therefore \;$ angle opposite to $s_3$ i.e. $S_3$ is the greatest angle.

By cosine rule,

$\begin{aligned} \cos \left(S_3\right) & = \dfrac{s_1^2 + s_2^2 - s_3^2}{2 s_1 s_2} \\\\ & = \dfrac{\dfrac{9a^2}{25} + a^2 - \dfrac{49 a^2}{25}}{2 \times \dfrac{3a}{5} \times a} \\\\ & = \dfrac{- 15 / 25}{6 / 5} \\\\ & = - \dfrac{1}{2} \end{aligned}$

$\implies$ $S_3 = \cos^{-1} \left(- \dfrac{1}{2}\right) = 120^\circ$

Properties of Triangles

If one angle of a triangle be $60^\circ$, the area $10 \sqrt{3} \; cm^2$ and the perimeter $20 \; cm$, find the lengths of the sides.

Let the sides of the triangle be $a$, $b$ and $c$.

Let $\;$ $A = 60^\circ$

Given: $\;$ Area of $\triangle ABC = \Delta = 10 \sqrt{3} \; cm^2$

Now, $\;$ $\Delta = \dfrac{1}{2}b c \sin A$

i.e. $\;$ $10 \sqrt{3} = \dfrac{1}{2}b c \sin \left(60^\circ\right)$

$\implies$ $bc = \dfrac{20 \sqrt{3}}{\sqrt{3} / 2}$ $\implies$ $bc = 40$ $\;\;\; \cdots \; (1)$

Given: $\;$ Perimeter of $\triangle ABC = 20 \; cm$

Perimeter of $\triangle ABC$ $= a + b + c$

$\therefore \;$ $a + b + c = 20$ $\implies$ $b + c = 20 - a$ $\;\;\; \cdots \; (2)$

By cosine rule, we have,

$\cos A = \dfrac{b^2 + c^2 - a^2}{2bc}$

$\begin{aligned} i.e. \; a^2 & = b^2 + c^2 - 2 b c \cos A \\\\ & = b^2 + c^2 - 2 b c \cos \left(60^\circ\right) \;\;\; \left[\because \; A = 60^\circ\right] \\\\ & = b^2 + c^2 - 2 \times b c \times \dfrac{1}{2} \\\\ & = b^2 + c^2 - bc \\\\ & = \left(b^2 + c^2 + 2 b c\right) - 3 b c \\\\ i.e. \; a^2 & = \left(b + c\right)^2 - 3 b c \;\;\; \cdots \; (3) \end{aligned}$

In view of equations $(1)$ and $(2)$, equation $(3)$ becomes,

$a^2 = \left(20 - a\right)^2 - 3 \times 40$

i.e. $\;$ $a^2 = 400 + a^2 - 40 a - 120$

i.e. $\;$ $40 a = 280$ $\implies$ $a = 7$

Substituting the value of $a$ in equation $(2)$ we get,

$b + c = 20 - 7 = 13$ $\implies$ $b = 13 - c$ $\;\;\; \cdots \; (4)$

In view of equation $(4)$, equation $(1)$ becomes,

$\left(13 - c\right)c = 40$

i.e. $\;$ $c^2 - 13 c + 40 = 0$

i.e. $\;$ $\left(c - 8\right) \left(c - 5\right) = 0$

$\implies$ $c = 8$ $\;$ or $\;$ $c = 5$

Substituting the value of $c$ in equation $(1)$ we get,

When $c = 8$, $\;$ $b = \dfrac{40}{c} = 5$

When $c = 5$, $\;$ $b = \dfrac{40}{c} = 8$

$\therefore \;$ The lengths of the sides of the triangle are:

$a = 7 \; cm$, $b = 5 \; cm$, $c = 8 \; cm$ $\;$ or $\;$ $a = 7 \; cm$, $b = 8 \; cm$, $c = 5 \; cm$

Properties of Triangles

A workman is told to make a triangular enclosure of sides $50$, $41$ and $21$ m respectively. Having made the first side one meter too long, what length must he make the other two sides in order to enclose the prescribed area with the prescribed length of fencing?

Actual lengths of sides: $\;$ $a = 50 \; m$, $\;$ $b = 41 \; m$, $\;$ $c = 21 \; m$

Actual perimeter $= a + b + c = 50 + 41 + 21 = 112 \; m$

Actual semi-perimeter $= s = \dfrac{a + b + c}{2} = 56 \; m$ $\;\;\; \cdots \; (1a)$

Actual area $= \Delta = \sqrt{s \left(s - a\right) \left(s - b\right) \left(s - c\right)}$

i.e. $\;$ $\Delta = \sqrt{56 \left(56 - 50\right) \left(56 - 41\right) \left(56 - 21\right)} = 420 \; m^2$ $\;\;\; \cdots \; (1b)$

Incorrect length $= a_1 = 51 \; m$

Let the remaining lengths be $b_1$ and $c_1$

New semi-perimeter $= s_1 = \dfrac{a_1 + b_1 + c_1}{2} = \dfrac{51 + b_1 + c_1}{2}$ $\;\;\; \cdots \; (2a)$

Given: $\;$ $s_1 = s$

$\therefore \;$ We have from equations $(1a)$ and $(2a)$

$\dfrac{51 + b_1 + c_1}{2} = 56$ $\implies$ $b_1 + c_1 = 61$ $\;\;\; \cdots \; (3)$

New Area $= \Delta_1 = \sqrt{s_1 \left(s_1 - a_1\right) \left(s_1 - b_1\right) \left(s_1 - c_1\right)}$

i.e. $\;$ $\Delta_1 = \sqrt{s \left(s - a_1\right) \left(s - b_1\right) \left(s - c_1\right)}$ $\;\;\;$ $\left[\because \; s_1 = s\right]$

Substituting the values of $s$ and $a_1$ we have,

$\Delta_1 = \sqrt{56 \left(56 - 51\right) \left(56 - b_1\right) \left(56 - c_1\right)}$

i.e. $\;$ $\Delta_1 = \sqrt{56 \times 5 \left(56 - b_1\right) \left(56 - c_1\right)}$ $\;\;\; \cdots \; (2b)$

Also given: $\;$ $\Delta_1 = \Delta$

$\therefore \;$ We have from equations $(1b)$ and $(2b)$

$\sqrt{56 \times 5 \left(56 - b_1\right) \left(56 - c_1\right)} = 420$

i.e. $\;$ $\sqrt{70 \left(56 - b_1\right) \left(56 - c_1\right)} = 210$

i.e. $\;$ $70 \left(56 - b_1\right) \left(56 - c_1\right) = \left(210\right)^2$

i.e. $\;$ $\left(56 - b_1\right) \left(56 - c_1\right) = 630$

i.e. $\;$ $3136 - 56 \left(b_1 + c_1\right) + b_1 c_1 = 630$

Substituting the value of $\left(b_1 + c_1\right)$ from equation $(3)$ we have,

$b_1 c_1 = 630 - 3136 + \left(56 \times 61\right) = 910$ $\;\;\; \cdots \; (4)$

Now, $\;$ $\left(b_1 - c_1\right)^2 = \left(b_1 + c_1\right)^2 - 4 b_1 c_1$ $\;\;\; \cdots \; (5)$

In view of equations $(3)$ and $(4)$, equation $(5)$ becomes

$\left(b_1 - c_1\right)^2 = \left(61\right)^2 - \left(4 \times 910\right) = 81$

$\implies$ $b_1 - c_1 = 9$ $\;\;\; \cdots \; (6)$

Adding equations $(5)$ and $(6)$ we have,

$2b_1 = 70$ $\implies$ $b_1 = 35$

Substituting the value of $b_1$ in equation $(3)$ gives, $c_1 = 61 - 35 = 26$

$\therefore \;$ The new lengths are $35 \; m$ and $26 \; m$

Properties of Triangles

Find the area of $\triangle ABC$ when $a = \sqrt{3}$, $b = \sqrt{2}$ and $c = \dfrac{\sqrt{6} + \sqrt{2}}{2}$

Given: $\;$ $a = \sqrt{3}$, $\;$ $b = \sqrt{2}$, $\;$ $c = \dfrac{\sqrt{6} + \sqrt{2}}{2}$

By cosine rule,

$\begin{aligned} \cos C & = \dfrac{a^2 + b^2 - c^2}{2 a b} \\\\ & = \dfrac{3 + 2 - \left(\dfrac{6 + 2 + 4 \sqrt{3}}{4}\right)}{2 \times \sqrt{3 \sqrt{2}}} \\\\ & = \dfrac{5 - \left(2 + \sqrt{3}\right)}{2 \sqrt{6}} \\\\ & = \dfrac{3 - \sqrt{3}}{2 \sqrt{6}} \\\\ & = \dfrac{\sqrt{3} - 1}{2 \sqrt{2}} \;\;\; \left[\text{dividing numerator and denominator by } \sqrt{3}\right] \end{aligned}$

$\implies$ $C = \cos^{-1} \left(\dfrac{\sqrt{3} - 1}{2 \sqrt{2}}\right) = 75^\circ$

Now, $\;$ area of $\triangle ABC$ $= \Delta = \dfrac{1}{2}a b \sin C$

$\begin{aligned} i.e. \; \Delta & = \dfrac{1}{2} \times \sqrt{3} \times \sqrt{2} \times \sin \left(75^\circ\right) \\\\ & = \dfrac{1}{2} \times \sqrt{6} \times \dfrac{\left(\sqrt{3} + 1\right)}{2 \sqrt{2}} \\\\ & = \dfrac{\sqrt{3} \left(\sqrt{3} + 1\right)}{4} \\\\ & = 1.183 \; \text{sq units} \end{aligned}$

Properties of Triangles

To get the distance of a point $A$ from a point $B$, a line $BC$ and the angles $ABC$ and $BCA$ are measured and found to be $287 \; cm$ and $55^\circ 32'10''$ and $51^\circ 8' 20''$ respectively. Find the distance $AB$.

Given: $\;$ $\angle ABC = B = 55^\circ 32' 10''$, $\;$ $\angle BCA = C = 51^\circ 8' 20''$, $\;$ $BC = 287 \; cm$

In $\triangle ABC$, $\;$ $A + B + C = 180^\circ$

$\begin{aligned} \therefore \; A & = 180^\circ - \left(B + C\right) \\\\ & = 180^\circ - \left(55^\circ 32' 10'' + 51^\circ 8' 20''\right) \\\\ & = 73^\circ 19' 30'' \end{aligned}$

In $\triangle ABC$, by sine rule, $\;$ $\dfrac{AB}{\sin C} = \dfrac{BC}{\sin A}$

$\therefore \;$ $AB = \dfrac{BC \; \sin C}{\sin A}$

$\begin{aligned} i.e. \; AB & = \dfrac{287 \times \sin \left(51^\circ 8' 20''\right)}{\sin \left(73^\circ 19' 30''\right)} \\\\ & = \dfrac{287 \times 0.7787}{0.9579} \\\\ & = 233.31 \; cm \end{aligned}$

Properties of Triangles

If the angles of a triangle be as $5 : 10 : 21$, and the side opposite the smaller angle be $3 \;$ cm, find the other sides.

Given: $\;$ $A : B : C = 5 : 10 : 21$

Let $k$ be the constant of proportionality.

Then, $\;$ $A = 5k$, $\;$ $B = 10 k$, $\;$ $C = 21 k$

In $\triangle ABC$, $\;$ $A + B + C = 180^\circ$

i.e. $\;$ $5 k + 10 k + 21 k = 180^\circ$

i.e $\;$ $36 k = 180^\circ$ $\implies$ $k = 5^\circ$

$\implies$ $A = 5 k = 25^\circ$, $\;$ $B = 10 k = 50^\circ$, $\;$ $C = 21 k = 105^\circ$

In $\triangle ABC$, by sine rule

$\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}$

i.e. $\;$ $b = \dfrac{a \; \sin B}{\sin A} = \dfrac{3 \times \sin \left(50^\circ\right)}{\sin \left(25^\circ\right)} = \dfrac{3 \times 0.7760}{0.4226} = 5.4378 \;$ cm

and $\;$ $c = \dfrac{a \; \sin C}{\sin A} = \dfrac{3 \times \sin \left(105^\circ\right)}{\sin \left(25^\circ\right)} = \dfrac{3 \times 0.9659}{0.4226} = 6.8568 \;$ cm

Properties of Triangles

The base angles of a triangle are $22.5^\circ$ and $112.5^\circ$. Prove that the base is equal to twice the height of the triangle.

Let the base angles be $B = 22.5^\circ$ and $C = 112.5^\circ$

Draw $AD \perp BC$ extended.

In $\triangle ABC$, $\;$ $A + B + C = 180^\circ$

$\therefore \;$ $A = 180^\circ - \left(B + C\right) = 180^\circ - \left(22.5^\circ + 112.5^\circ\right) = 45^\circ$

In $\triangle ABC$, by sine rule,

$\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C} = k$ (say), where $k$ is the constant of proportionality.

$\implies$ $b = \dfrac{a \sin B}{\sin A} = \dfrac{a \sin \left(22.5^\circ\right)}{\sin \left(45^\circ\right)}$ $\;\;\; \cdots \; (1)$

Now, in right $\triangle ACD$,

$\begin{aligned} AD & = AC \; \sin \left(\angle ACD\right) \\\\ & = b \; \sin \left(180^\circ - \angle ACB\right) \\\\ & = b \; \sin \left(\angle ACB\right) \\\\ & = b \; \sin \left(112.5^\circ\right) \\\\ & = \dfrac{a \times \sin \left(22.5^\circ\right) \times \sin \left(112.5^\circ\right)}{\sin \left(45^\circ\right)} \;\; \left[\text{by equation (1)}\right] \\\\ & = \dfrac{a \times \left(\dfrac{\sqrt{2 - \sqrt{2}}}{2}\right) \times \left(\dfrac{\sqrt{2 + \sqrt{2}}}{2}\right)}{\dfrac{1}{\sqrt{2}}} \\\\ & = \dfrac{\sqrt{2} \times a \times \sqrt{2}}{4} \\\\ & = \dfrac{a}{2} \end{aligned}$

$\implies$ $a = 2 \; AD$

i.e. $\;$ The base of $\triangle ABC$ is equal to twice its height.

Hence proved.