Trigonometry - Heights and Distances

A straight road leads to the foot of a tower $200 \; m$ high. From the top of the tower, the angles of depression of two cars standing on the road are observed to be $45^\circ$ and $60^\circ$ respectively. Find the distance between the two cars. Give your answer correct to two decimal places. Take $\sqrt{3} = 1.732$.

$OT$: Tower of height $200 \; m$

$C_1, \; C_2$: Positions of the two cars

$C_1C_2$: Distance between the two cars

In $\triangle OTC_1$,

$\dfrac{OT}{OC_1} = \tan 60^\circ = \sqrt{3}$

i.e. $\;$ $OC_1 = \dfrac{OT}{\sqrt{3}} = \dfrac{200}{\sqrt{3}} \; m$ $\;\;\; \cdots \; (1)$

In $\triangle OTC_2$,

$\dfrac{OT}{OC_2} = \tan 45^\circ = 1$

i.e. $\;$ $OC_2 = OT = 200 \; m$ $\;\;\; \cdots \; (2)$

Now, $\;$ $C_1 C_2 = OC_2 - OC1$

i.e. $\;$ $C_1 C_2 = 200 - \dfrac{200}{\sqrt{3}}$ $\;\;\;$ [in view of equations $(1)$ and $(2)$]

i.e. $\;$ $C_1 C_2 = 200 \left(\dfrac{\sqrt{3} - 1}{\sqrt{3}}\right) = 84.53 \; m$ $\;\;\;$ [correct to two decimal places]

Trigonometry - Heights and Distances

An airplane at a height of $6 \; km$ passes vertically above another plane at an instant when their angles of elevation at the same observing point are $60^\circ$ and $45^\circ$ respectively. Find the difference between the heights of the planes. Take $\sqrt{3} = 1.732$

$P_1, \; P_2 =$ Airplanes

$O =$ Point of observation

$AP_1 =$ Height of airplane $P_1 = 6 \; km$

$P_1 P_2 =$ Difference between the heights of the planes

In $\triangle OP_1A$, $\;$ $\dfrac{AP_1}{OA} = \tan 60^\circ$

$\implies$ $OA = \dfrac{AP_1}{\tan 60^\circ} = \dfrac{6}{\sqrt{3}} = 2\sqrt{3} \; km$

In $\triangle OP_2A$, $\;$ $\dfrac{AP_2}{OA} = \tan 45^\circ$

$\implies$ $AP_2 = OA \times \tan 45^\circ = 2 \sqrt{3} \times 1 = 2 \sqrt{3} \; km$

Difference between the heights of the planes $= P_1P_2$

$P_1 P_2 = AP_1 - AP_2 = 6 - 2 \sqrt{3} = 2.536 \; km$

Trigonometry

The horizontal distance between two towers is $150 \; m$. The angle of depression of the top of one tower as observed from the top of the second tower, which is $120 \; m$ in height, is $30^\circ$. Find the height of the first tower. Take $\sqrt{3} = 1.732$.

$O_1 T_1, \; O_2 T_2 = $ two towers

Height of tower $O_1 T_1 = 120 \; m$

Height of tower $O_2 T_2 = h$ meter

Distance between the two towers $= O_1 O_2 = 150 \; m$

From the figure, $\;$ $XT_2 = O_1 O_2 = 150 \; m$

In $\triangle T_1 X T_2$, $\;$ $\tan 30^\circ = \dfrac{X T_1}{X T_2}$

$\implies$ $X T_1 = X T_2 \times \tan 30^\circ = 150 \times \dfrac{1}{\sqrt{3}} = 50 \sqrt{3} \; m$

But $\;$ $X T_1 = O_1 T_1 - O_2 T_2$

i.e. $\;$ $50 \sqrt{3} = 120 - h$

$\implies$ $h = 120 - 50 \sqrt{3} = 120 - 86.6 = 33.4 \;m$

$\therefore \;$ Height of tower $O_2 T_2 = 33.4 \; m$

Trigonometry

As observed from the top of a $80 \; m$ tall light house, the angles of depression of two ships on the same side of the lighthouse, in horizontal line with its base, are $30^\circ$ and $40^\circ$
Find the distance between the two ships, giving the answer correct to the nearest meter.

$OL: $ Light house of height $= 80 \; m$

$S_1, \; S_2: $ Two ships

$S_1 S_2 = $ Distance between the ships

In $\triangle OLS_1$, $\;$ $\dfrac{OL}{OS_1} = \tan 40^\circ$

i.e. $\;$ $OS_1 = \dfrac{OL}{\tan 40^\circ}$ $\;\;\; \cdots \; (1)$

In $\triangle PLS_2$, $\;$ $\dfrac{OL}{OS_2} = \tan 30^\circ$

i.e. $\;$ $OS_2 = \dfrac{OL}{\tan 30^\circ}$

i.e. $\;$ $OS_1 + S_1 S_2 = \dfrac{OL}{\tan 30^\circ}$ $\;\;\; \cdots \; (2)$

In view of equation $(1)$, equation $(2)$ becomes

$\dfrac{OL}{\tan 40^\circ} + S_1 S_2 = \dfrac{OL}{\tan 30^\circ}$

i.e. $\;$ $S_1 S_2 = OL \left[\dfrac{1}{\tan 30^\circ} - \dfrac{1}{\tan 40^\circ}\right]$

i.e. $\;$ $S_1 S_2 = 80 \left[\dfrac{1}{0.5774} - \dfrac{1}{0.8391}\right] = 43.21$

$\therefore \;$ Distance between the ships $= S_1 S_2 = 43 \; m$ (to the nearest meter)

Trigonometry

A tower subtends an angle $\alpha$ on the same level as the foot of the tower and at a second point $h$ meters above the first, the depression of the foot of the tower is $\beta$. Show that the height of the tower is $h \tan \alpha \cot \beta$.

In the figure,

$AT = $ tower of height $H$

$B = $ a point on the same level as the foot of the tower at a distance $x$ from $A$

$C = $ a point $h$ meters above $B$

From the figure,

in $\triangle ATB$, $\;$ $\tan \alpha = \dfrac{AT}{AB} = \dfrac{H}{x}$

in $\triangle ACB$, $\;$ $\tan \beta = \dfrac{BC}{AB} = \dfrac{h}{x}$

$\therefore \;$ $\dfrac{\tan \alpha}{\tan \beta} = \dfrac{H / x}{h / x} = \dfrac{H}{h}$

i.e. $\;$ $H = \dfrac{\tan \alpha}{\tan \beta} \times h$

i.e. $\;$ $H = h \tan \alpha \cot \beta$

Trigonometry

From the top of a building $60 \; m$ high, angles of elevation and depression were observed to the top and base of a tower to be $30^\circ$ and $60^\circ$ respectively. Find the height of the tower, given both the structures are on the same level ground and opposite each other.

$AB = $ Building of height $60 \; m$

$OT = $ Tower

$AO = $ Distance between the building and the tower

From $B$ draw $BP \perp OT$

Then, $\;$ $AB = OP = 60 \; m$ $\;$ and $\;$ $AO= BP$

In $\triangle ABO$, $\;$ $\dfrac{AB}{AO} = \tan 60^\circ$

$\implies$ $AO = \dfrac{AB}{\tan 60^\circ} = \dfrac{60}{\sqrt{3}} = 20 \sqrt{3} \; m$

$\therefore \;$ $BP = AO = 20 \sqrt{3} \; m$

In $\triangle PTB$, $\;$ $\dfrac{PT}{BP} = \tan 30^\circ$

$\implies$ $PT = BP \times \tan 30^\circ = 20 \sqrt{3} \times \dfrac{1}{\sqrt{3}} = 20 \; m$

Height of tower $= OP + PT = 60 + 20 = 80 \; m$

Trigonometry

The angle of elevation, observed from ground to the top of a building $60 \; m$ high, is $30^\circ$. On walking '$x$' meters towards the building, the angle of elevation to the top changes to $60^\circ$. Find '$x$' to the nearest meter.

$AB =$ Building of height $60 \; m$

$D = $ Initial position of observation

$C = $ Final position of observation

$CD = x \;$ meter

In $\triangle ABC$,

$\tan 60^\circ = \dfrac{AB}{BC}$

i.e. $\;$ $BC = \dfrac{60}{\sqrt{3}}$ $\;\;\; \cdots \; (1)$

In $\triangle ABD$,

$\tan30^\circ = \dfrac{AB}{BD} = \dfrac{AB}{BC + CD} = \dfrac{AB}{BC + x}$

i.e. $\;$ $BC + x = \dfrac{60}{1 / \sqrt{3}} = 60 \sqrt{3}$

i.e. $\; $ $x = 60 \sqrt{3} - BC$ $\;\;\; \cdots \; (2)$

In view of equation $(1)$ equation $(2)$ becomes,

$x = 60 \sqrt{3} - \dfrac{60}{\sqrt{3}} = \dfrac{180 - 60}{\sqrt{3}} = \dfrac{120}{\sqrt{3}}= 40 \sqrt{3} = 69.28 \;$ meter

i.e. $\;$ $x = 69 \; m$ (to the nearest meter)

Trigonometry

A man in a boat rowing away from a lighthouse $150 \; m$ high takes $1$ minute to change the angle of elevation of the top of the lighthouse from $60^\circ$ to $45^\circ$. Find the speed of the boat in meter per minute. [Given: $\sqrt{3} = 1.732$]

$AB =$ Lighthouse of height $150 \; m$

$C =$ First point of observation

$D =$ Second point of observation

$CD =$ Distance traveled by the boat in $1$ minute

In $\triangle ABD$, $\;$ $\dfrac{AB}{BD} = \tan 45^\circ = 1$

$\implies$ $BD = AB = 150 \; m$

In $\triangle ABC$, $\;$ $\dfrac{AB}{BC} = \tan 60^\circ = \sqrt{3}$

$\implies$ $BC = \dfrac{AB}{\sqrt{3}} = \dfrac{150}{\sqrt{3}} = 50 \sqrt{3} = 86.6 \; m$

Now, $\;$ $CD = BD - BC = 150 - 86.6 = 63.4 \; m$

$\therefore \;$ Distance covered by the boat in $1$ minute $= 63.4 \; m$

$\therefore \;$ Speed of the boat $= 63.4 \; m/\text{min}$

Heights and Distances

A building and a statue are in opposite side of a street from each other $35 \; m$ apart. From a point on the roof of the building, the angle of elevation of the top of statue is $24^\circ$ and the angle of depression of the base of the statue is $34^\circ$. Find the height of the statue. Give your answer correct to two decimal places. [Given: $\tan 24^\circ = 0.4452$, $\tan 34^\circ = 0.6745$]

$AB =$ Building

$ST =$ Statue

$AS = 35 \; m = $ Distance between the building and the statue

Draw $BO \perp ST$.

Then, $\;$ $AB = SO$, $\;$ $AS = BO = 35 \; m$

In $\triangle ABS$, $\;$ $\dfrac{AB}{AS} = \tan 34^\circ$

$\implies$ $AB = AS \tan 34^\circ = 35 \times 0.6745 = 23.6075 \; m$

i.e. $\;$ $SO = 23.6075 \; m$

In $\triangle BTO$, $\;$ $\dfrac{OT}{BO} = \tan 24^\circ$

$\implies$ $OT = BO \tan 24^\circ = 35 \times 0.4452 = 15.582 \; m$

Now, $\;$ $ST = SO + OT = 23.6075 + 15.582 = 39.1895 \; m$

$\therefore \;$ The height of the statue $= 39.19 \; m$ (to 2 decimal places)

Heights and Distances

An airplane is flying parallel to the Earth’s surface at a speed of $175 \; m/s$ and at a height of $600 \; m$. The angle of elevation of the airplane from a point on the Earth’s surface is $37^\circ$. After what period of time does the angle of elevation increase to $53^\circ$? Give your answer correct to two decimal places. [Given: $\tan 53^\circ = 1.3270$, $\tan 37^\circ = 0.7536$]

$A_1 =$ Initial position of the airplane from point $O$

$A_2 =$ Final position of the airplane from point $O$

$A_1 P = A_2 Q = 600 \; m =$ Height of the airplane

In $\triangle OA_1P$, $\;$ $\dfrac{A_1P}{OP} = \tan 37^\circ$

$\implies$ $OP = \dfrac{A_1 P}{\tan 37^\circ} = \dfrac{600}{0.7536} = 796.1783 \; m$

In $\triangle OA_2Q$, $\;$ $\dfrac{A_2Q}{OQ} = \tan 53^\circ$

$\implies$ $OQ = \dfrac{A_2 Q}{\tan 53^\circ} = \dfrac{600}{1.3270} = 452.1477 \; m$

Now, $\;$ $A_1 A_2 = OP - OQ = 796.1783 - 452.1477 = 344.0306 \; m$

Given: Speed of plane $= 175 \; m/s$

i.e. $\;$ a distance of $175 \; m$ is traveled in $1 \; s$

$\therefore \;$ a distance of $344.0306 \; m$ is traveled in $\dfrac{344.0306}{175} = 1.9659 \; s$

$\therefore \;$ The time taken for the plane to go from $A_1$ to $A_2$ is $1.97 \; s$ (to 2 decimal places)

Heights and Distances

Two persons are standing '$x$' meter apart from each other. The height of the first person is double that of the other. If from the middle point of the line joining their feet, an observer finds the angular elevations of their tops to be complementary, find the height of the shorter person.

$P_1, \; P_2 =$ Two persons standing at $O$ and $Q$ respectively

$OP_1 =$ Height of $P_1 = 2h$ meter

$OP_2 =$ Height of $P_2 = h$ meter

$M =$ Midpoint between $O$ and $Q$

Let $\;$ $\angle P_1MO = \theta$

Then, $\;$ $\angle P_2MQ = 90 - \theta$

$OQ =$ Distance between the two people $= x$ meter

In $\triangle MP_2Q$, $\;$ $\dfrac{P_2Q}{MQ} = \tan \left(90 - \theta\right) = \cot \theta$

i.e. $\;$ $\cot \theta = \dfrac{h}{x / 2}$ $\;\;\; \cdots \; (1)$

In $\triangle MPO$, $\;$ $\dfrac{PO}{OM} = \tan \theta$

i.e. $\;$ $\tan \theta = \dfrac{2h}{x/2}$ $\;\;\; \cdots \; (2)$

Now, $\;$ $\tan \theta \times \cot \theta = 1$ $\;\;\; \cdots \; (3)$

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

$\dfrac{2h}{x / 2} \times \dfrac{h}{x / 2} = 1$

i.e. $\;$ $8h^2 = x^2$ $\implies$ $h = \dfrac{x}{2 \sqrt{2}}$

$\therefore \;$ The height of the shorter person $= \dfrac{x}{2 \sqrt{2}}$ meter

Heights and Distances

The angle of depression of the top and bottom of a $20 \; m$ tall building from the top of a multi-storied building are $30^\circ$ and $60^\circ$ respectively. Find the height of the multi-storied building and the distance between two buildings.

$AB =$ Multi-storied building

$CD =$ Building of height $20 \; m$

$AC =$ Distance between the two buildings

Draw $DP \perp AB$

Then $\;$ $AP = CD = 20$ and $PD = AC$

In $\triangle BPD$, $\;$ $\dfrac{PB}{PD} = \tan 30^\circ$

$\implies$ $PB = PD \; \tan 30^\circ = \dfrac{PD}{\sqrt{3}} = \dfrac{AC}{\sqrt{3}}$ $\;\;\; \cdots \; (1)$

In $\triangle BAC$, $\;$ $\dfrac{AB}{AC} = \tan 60^\circ$

$\implies$ $AB = AC \; \tan 60^\circ = \sqrt{3} \; AC$ $\;\;\; \cdots \; (2)$

But $\;$ $AB = AP + PB$ $\;\;\; \cdots \; (3)$

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

$\sqrt{3} \; AC = 20 + \dfrac{AC}{\sqrt{3}}$

i.e. $\;$ $AC \left(\sqrt{3} - \dfrac{1}{\sqrt{3}}\right) = 20$

i.e. $\;$ $\dfrac{2 \; AC}{\sqrt{3}} = 20$ $\implies$ $AC = 10 \sqrt{3}$

Substituting the value of $AC$ in equation $(2)$ we have,

$AB = 10 \sqrt{3} \times \sqrt{3} = 30$

$\therefore \;$ Height of the multi-storied building $= 30 \; m$

Distance between the two buildings $= 10 \sqrt{3} \; m$

Heights and Distances

Three people $A$, $B$ and $C$ can see each other using telescope across a valley. The horizontal distance between $A$ and $B$ is $8 \; km$ and the horizontal distance between $B$ and $C$ is $12 \; km$. The angle of depression of $B$ from $A$ is $20^\circ$ and the angle of elevation of $C$ from $B$ is $30^\circ$. Calculate the vertical height between $A$, $B$ and $B$, $C$. [Given: $\tan 20^\circ = 0.3640$, $\sqrt{3} = 1.732$]

$A'B = 8 \; km =$ Horizontal distance between $A$ and $B$

$BC' = 12 \; km =$ Horizontal distance between $B$ and $C$

$AA' =$ Vertical height between $A$ and $B$

$CC' =$ Vertical height between $B$ and $C$

In $\triangle AA'B$, $\;$ $\dfrac{AA'}{AB} = \tan 20^\circ$

$\therefore \;$ $AA' = AB \; \tan 20^\circ = 8 \times 0.3640 = 2.912$

In $\triangle CC'B$, $\;$ $\dfrac{CC'}{BC'} = \tan 30^\circ$

$\implies$ $CC' = BC' \; \tan 30^\circ = 12 \times \dfrac{1}{\sqrt{3}} = 4 \sqrt{3} 4 \times 1.732 = 6.925$

$\therefore \;$ Vertical height between $A$ and $B$ $= 2.912 \; km$

Vertical height between $B$ and $C$ $= 6.925 \; km$

Heights and Distances

A man is standing on the deck of a ship, which is $40 \; m$ above water level. He observes the angle of elevation of the top of a hill as $60^\circ$ and the angle of depression of the base of the hill as $30^\circ$. Calculate the distance of the hill from the ship and the height of the hill. [Given: $\sqrt{3} = 1.732$]

$M =$ Position of man on the deck of a ship

$O = $ Position of the ship

$MO = 40 \; m$ Height of the deck of ship above water level

$H =$ Position of the hill

$HH' =$ Height of the hill

$OH =$ Distance of the hill from the ship

Draw $MP \perp HH'$

Then, $\;$ $MP = OH$, $MO = PH = 40 \; m$

In $\triangle OMH$, $\;$ $\dfrac{OM}{OH} = \tan 30^\circ = \dfrac{1}{\sqrt{3}}$

$\implies$ $OH = OM \times \sqrt{3} = 40 \times \sqrt{3} = 40 \times 1.732 = 69.28$

In $\triangle MPH'$, $\;$ $\dfrac{H'P}{MP} = \tan 60^\circ = \sqrt{3}$

$\implies$ $H'P = MP \times \sqrt{3} = 40 \sqrt{3} \times \sqrt{3} = 120$

Now, $HH' = H'P + PH = 120 + 40 = 160$

$\therefore \;$ Distance of the hill from the ship $= 69.28 \; m$

Height of hill $= 160 \; m$

Heights and Distances

From the top of a lighthouse, the angle of depression of two ships on the opposite sides of it are observed to be $30^\circ$ and $60^\circ$. If the height of the lighthouse is $h$ meter and the line joining the ships passes through the foot of the lighthouse, show that the distance between the ships is $\dfrac{4h}{\sqrt{3}} \; m$

$OL =$ Lighthouse of height $h$ meter

$S_1, \; S_2 =$ Position of the two ships on either side of the lighthouse

$S_1S_2 =$ Distance between the two ships

In $\triangle S_1OL$, $\;$ $\dfrac{OL}{S_1O} = \tan 30^\circ$

i.e. $\;$ $\dfrac{h}{S_1O} = \dfrac{1}{\sqrt{3}}$

$\implies$ $S_1O = h \sqrt{3}$

In $\triangle S_2OL$, $\;$ $\dfrac{OL}{OS_2} = \tan 60^\circ$

i.e. $\;$ $\dfrac{h}{OS_2} = \sqrt{3}$

$\implies$ $OS_2 = \dfrac{h}{\sqrt{3}}$

Now, $\;$ $S_1S_2 = S_O + OS_2 = h \sqrt{3} + \dfrac{h}{\sqrt{3}} = \dfrac{4h}{\sqrt{3}}$

$\therefore \;$ Distance between the two ships $= \dfrac{4h}{\sqrt{3}}$ meter

Heights and Distances

The horizontal distance between two buildings is $70 \; m$. The angle of depression of the top of the first building when seen from the top of the second building is $45^\circ$. If the height of the second building is $120 \; m$, find the height of the first building.

$AA'$, $BB'$ are two buildings

$AB = $ distance between the two buildings $= 70 \; m$

$BB' =$ building of height $120 \; m$

Let height of building $AA' = h$ meter

Draw $A'P \perp BB'$.

Then $A'P = AB = 70 \;m$ and $AA' = BP$.

In $\triangle PB'A'$, $\;$ $\dfrac{PB'}{A'P} = \tan 45^\circ = 1$

$\implies$ $PB' = A'P = 70 \; m$

Now, $\;$ $BB' = BP + PB'$

i.e. $\;$ $BB' = AA' + PB'$

$\therefore \;$ $AA' = h = BB' - PB' = 120 - 70 = 50 \; m$

$\therefore \;$ Height of the building $= 50 \; m$

Heights and Distances

A flag pole of height '$h$' meters is on top of a hemispherical dome of radius '$r$' meters. A man standing $7 \; m$ away from the dome sees the top of the pole at an angle of $45^\circ$. On moving $5 \; m$ away from the dome, he sees the bottom of the pole at an angle of $30^\circ$. Find the height of the pole and the radius of the dome. [Given: $\sqrt{3} = 1.732$]

$OO' = OA = r = $ Radius of the dome

$O'P = $ flag pole of height $h$ meter

$B = $ Initial point of observation such that $AB = 7 \; m$

$C = $ Final point of observation such that $BC = 5 \; m$

In $\triangle OO'C$, $\;$ $\dfrac{OO'}{OC} = \tan 30^\circ = \dfrac{1}{\sqrt{3}}$

$\implies$ $OO' = \dfrac{OC}{\sqrt{3}}$

i.e. $\;$ $OO' = \dfrac{OA + AB + BC}{\sqrt{3}} = \dfrac{OA + 12}{\sqrt{3}}$

i.e. $\;$ $OO' \times \sqrt{3} = OA + 12$

i.e. $\;$ $OO' \times \sqrt{3} = OO' + 12$

i.e. $\;$ $\left(\sqrt{3} - 1\right)OO' = 12$

$\implies$ $OO' = \dfrac{12}{0.732} = 16.39$

In $\triangle OPB$, $\;$ $\dfrac{OP}{OB} = \tan 45^\circ = 1$

$\implies$ $OP = OB$

i.e. $\;$ $h + r = r + AB$

i.e. $\;$ $h = AB = 7$

$\therefore \;$ Height of pole $= 7 \; m$

Radius of dome $= 16.39 \; m$

Heights and Distances

A statue $1.6 \; m$ tall stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is $60^\circ$ and from the same point the angle of elevation of the top of the pedestal is $40^\circ$. Find the height of the pedestal. [Given: $\tan 40^\circ = 0.8391, \; \sqrt{3} = 1.732$]

$PS =$ Statue of height $1.6 \; m$

$OP =$ Pedestal

$A =$ Point of observation

In $\triangle SOA$, $\;$ $\dfrac{SO}{OA} = \tan 60^\circ = \sqrt{3}$

$\implies$ $OA = \dfrac{SO}{\sqrt{3}} = \dfrac{SP + PO}{\sqrt{3}}$

i.e. $\;$ $OA = \dfrac{1.6 + PO}{1.732}$ $\;\;\; \cdots \; (1)$

In $\triangle POA$, $\;$ $\dfrac{PO}{OA} = \tan 40^\circ = 0.8391$

$\implies$ $OA = \dfrac{PO}{0.8391}$ $\;\;\; \cdots \; (2)$

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

$\dfrac{1.6 + PO}{1.732} = \dfrac{PO}{0.8391}$

i.e. $\;$ $1.34256 + 0.8391 \; PO = 1.732 \; PO$

i.e. $\;$ $0.8929 \; PO = 1.34256$ $\implies$ $PO = 1.5036$

$\therefore \;$ Height of the pedestal $= 1.5 \; m$

Heights and Distances

To a man standing outside his house, the angles of elevation of the top and bottom of a window are $60^\circ$ and $45^\circ$ respectively. If the height of the man is $180 \; cm$ and if he is $5 \; m$ away from the wall, what is the height of the window and the wall? [Given: $\sqrt{3} = 1.732$]

$W_1$: Top of window

$W_2$: Bottom of window

$OW_2$: Wall

$AB$: Man of height $180 \; cm = 1.8 \; m$

$OA$: Distance of man from wall $= 5 \; m$

Draw $BB' \perp O W_2 W_1$

In the figure, $\;$ $AB = OB' = 1.8 \; m$ $\;$ and $\;$ $BB' = OA = 5 \; m$

In $\triangle W_2B'B$, $\;$ $\dfrac{W_2B'}{BB'} = \tan 45^\circ = 1$

$\implies$ $W_2 B' = BB' = 5 \; m$

In $\triangle W_1B'B$, $\;$ $\dfrac{W_1B'}{BB'} = \tan 60^\circ = \sqrt{3}$

$\implies$ $W_1B' = BB' \sqrt{3} = 5 \sqrt{3} \; m$

$\therefore \;$ Height of window $= W_1W_2 = W_1 B' - W_2 B' = 5 \left(\sqrt{3} - 1\right) = 5 \times 0.732 = 3.66 \; m$

Height of wall $= W_2 O = W_2 B' + B'O = 5 + 1.8 = 6.8 \; m$

Heights and Distances

An airplane is flying at a height of $300 \; m$ above the ground. Flying at this height, the angles of depression from the airplane of two points on both banks of a river in opposite directions are $45^\circ$ and $60^\circ$ respectively. Find the width of the river. [Given: $\sqrt{3} = 1.732$]

$B_1$, $B_2$: Points on opposite sides of the river bank

$P$: Position of the airplane

$PO$: Height of plane $= 300 \; m$

$B_1B_2$: Width of the river

In $\triangle POB_1$, $\;$ $\dfrac{PO}{OB_1} = \tan 60^\circ = \sqrt{3}$

$\therefore \;$ $OB_1 = \dfrac{PO}{\sqrt{3}} = \dfrac{300}{\sqrt{3}} = 100 \sqrt{3} = 100 \times 1.732 = 173.2 \; m$

In $\triangle POB_2$, $\;$ $\dfrac{PO}{OB_2} = \tan 45^\circ = 1$

$\implies$ $OB_2 = PO = 300 \; m$

$\therefore \;$ Width of river $= OB_1 + OB_2 = 173.2 + 300 = 473.2 \; m$