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.

Properties of Triangles

If $A = 45^\circ$, $B = 75^\circ$ and $C = 60^\circ$, prove that $a + c \sqrt{2} = 2b$

Given: $\;$ $A = 45^\circ$, $\;$ $B = 75^\circ$, $\;$ $C = 60^\circ$

By sine rule,

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

where $k$ is a constant of proportionality.

$\therefore \;$ We have, $\;$ $a = k \sin A = k \sin 45^\circ = \dfrac{k}{\sqrt{2}}$ $\;\;\; \cdots \; (1a)$

$b = k \sin B = k \sin 75^\circ = \dfrac{k \left(\sqrt{3} + 1\right)}{2 \sqrt{2}}$ $\;\;\; \cdots \; (1b)$

$c = k \sin C = k \sin 60^\circ = \dfrac{\sqrt{3} k}{2}$ $\;\;\; \cdots \; (1c)$

Now, from equations $(1a)$ and $(1c)$

$\begin{aligned} a + c \sqrt{2} & = \dfrac{k}{\sqrt{2}} + \dfrac{\sqrt{3} \times \sqrt{2} \; k}{2} \\\\ & = \dfrac{2 k + 2 \sqrt{3} \; k}{2 \sqrt{2}} \\\\ & = \dfrac{2 k \left(\sqrt{3} + 1\right)}{2 \sqrt{2}} \\\\ & = 2b \;\;\; \left[\text{in view of equation (1b)}\right] \end{aligned}$

$\therefore \;$ $a + c \sqrt{2} = 2 b$

Hence proved.