 # Some Applications of Trigonometry

## Class 10 NCERT

### NCERT

1   A circus artist is climbing a 20 m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground level is 30 °. ##### Solution :

It can be observed from the figure that AB is the pole.\\ $In \Delta{ABC}, \\ \frac{AB}{ AC} = \sin30^o \\ \frac{AB}{20} = \frac{1}{2} \\ AB= \frac{20}{2} = 10 \\ Therefore the height of the pole is 10 m.$

2   A tree breaks due to storm and broken part bends so that the top of the tree touches the ground making an angle $30^o$ with it. The distance between the foot of the tree to the point where the top touches the ground is $8 m$. Find the height of the tree.

##### Solution : Let AC was a original tree. Due to storm, it was broken into the parts.The broken parts A'B is making $30^o$ with the ground. $In \Delta A'BC, \\ \dfrac{BC}{A'C} = \tan30^o \\ \dfrac{BC}{8} = \dfrac{ 8}{ \sqrt{3}} \\ BC=\left(\dfrac{8}{\sqrt{3}}\right)m \\ \dfrac{A'C}{A'B} = \cos 30^o \\ \dfrac{8}{A'B} = \dfrac{\sqrt{3}}{2} \\ A'B = \left(\dfrac{16}{\sqrt{3}}\right)m \\ Height of tree = A'B +BC \\ = \left(\dfrac{16}{\sqrt{3}} + \dfrac{8}{\sqrt{3}}\right) = \dfrac{24}{\sqrt{3}}m \\ = 8\sqrt{3}m$

3   A contractor plans to install two slides for the children to play i a park. For the children below the age of $5$ years, she prefers to have a slide whose top is at a height of $1.5m$, and is inclined at an angle of $30^o$ to the ground, whereas for the elder children she wants to have a steep slide at a height of $3m$, and inclind at an angle of $60^o$ to the ground. what should be the length of the slide in each case?

##### Solution : It can be observed that AC and PR are the slides for younger and elder children respectively. $In \Delta ABC, \\ \\ \dfrac{AB}{AC} = \sin30^o \\ \dfrac{1.5}{AC} = \dfrac{1}{2} \\ AC = 3m \\ \ \ \ \ \ \ \ \ \ \\$ $In \Delta PQR, \\ \dfrac{PQ}{PR} = \sin60^o \\ \dfrac{3}{PR} = \dfrac{\sqrt{3}}{2} \\ PR = \dfrac{6}{\sqrt{3}} = \dfrac{2}{\sqrt{3}}m \\ Therefore, the lengths of these slides are 3m and 2\sqrt{3}m.$

4   The angle of elevation of the top of a tower from a point on the ground, which is $30m$ away from the foot of the tower is $30^o$.Find the height of the tower.

##### Solution : Let AB be the tower and the angle of elevation from point C(on ground) is $30^o$. $In 'ABC, \\ \dfrac{AB}{BC} = \tan30^o \\ \dfrac{AB}{30} = \dfrac{1}{\sqrt{3}} \\ AB = \dfrac{30}{\sqrt{3}} = 10{\sqrt{3}}m \\ Therefore the height of the tower is 10\sqrt{3}m.$

5   A kite is flying at a height of $60m$ abov the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is $60^o$. Find the length of the string, assuming that there is no slack in the string.

##### Solution : Let $k$ be the kite and the string is tied to point P on the ground. $In \Delta KLP, \\ \dfrac{KL}{KP} = \sin60^o \\ \dfrac{60}{KP} = \dfrac{\sqrt{3}}{2} \\ KP = \dfrac{120}{\sqrt{3}} = 40\sqrt{3}m \\ Hence,the length of the string is 40\sqrt{3}m.$

6   A $1.5m$ tall boy is standing at some distance from a $30 m$ tall building. The angle of elevation from his eyes to the top of the building increases from $30^o$ to $60^o$ as he walks towards the building. Find the distance he walked towards the building.

##### Solution : Let the boy was standing at point S initially.He walked towards the building and reached at point $T$. It can be observed that $PR = PQ-RQ \\ = \left({30-1.5}\right)m = 28.5m = \dfrac{57}{2}m \\ \ \ \ \ \\ In \Delta PAR, \\ \dfrac{PR}{AR} = \tan30^o \\ \dfrac{57}{2AR} = \dfrac{1}{\sqrt{3}} \\ AR = \left({\dfrac{57*\sqrt{3}}{2}}\right)m \\ In \Delta PRB, \\ \dfrac{PR}{BR} = \tan60^o \\ \dfrac{57}{2BR} = \sqrt{3} \\ BR = \dfrac{57}{2 \sqrt{3}} = \left( \dfrac{19 \sqrt{3}}{2} \right)m \\ \ \ \ \ \ \ \\ ST = AB \\ = AR-BR = \left( \dfrac{57 \sqrt{3}}{2} - \dfrac{19 \sqrt{3}}{2}m \right) = \left( \dfrac{38 \sqrt{3}}{2} \right)m = 19 \sqrt{3}m$ Hence, he walked $19\sqrt{3}m$ towards the building.

7   From a point on the ground, the angles of the bottom and the top of a transmission tower fixed at the top of a $20 m$ high building are $45^o$ and $60^o$ respectively. Find the height of the tower.

##### Solution : Let BC be the building,AB be the transmission tower, and D be the point on the ground from where the elevation angles are to be measured. $In \Delta BCD, \\ \dfrac{BC}{CD} = \tan45^o \dfrac{20}{CD} = 1 \\ CD = 20m \\ \ \ \ \ \ In \Delta ACD, \\ \dfrac{AC}{CD} = \tan60^o \\ \dfrac{AB+BC}{CD} = \sqrt{3} \\ \dfrac{AB+20}{CD} = \sqrt{3} \\ AB=\left(20\sqrt{3} - 20\right)m\\ = 20\left(\sqrt{3} -1\right)m \\$ Therefore, the height of the transmission tower is $20\left(\sqrt{3} - 1 \right)m$.

8   A statue $1.6m$ tall, stands on a top of pedestal,from a point on the ground, the angle of elevation of the top of statue is $60^o$ and from the same point the angle of elevation of the top of the pedestal is $45^o$ . Find the height of the pedestal.

##### Solution : Let AB the statue, BC be the pedestal, and D be the point on the ground from where the elevation angles are to be measured. $In \Delta BCD, \\ \dfrac{BC}{CD} = \tan45^o \\ \dfrac{BC}{CD} = 1\\ BC=CD\\ \ \ \ \ \ \\ In \Delta ACD, \\ \dfrac{AB+BC}{CD} = \tan60^o \\ \dfrac{AB+BC}{CD} = \sqrt{3} \\ 1.6 +BC =BC\sqrt{3} \ \ \ \ \left[As CD = BC \right] \\ BC\left(\sqrt{3} - 1\right) = 1.6 \\ BC = \dfrac{\left(1.6\right)\left(\sqrt{3}+1\right)}{\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)} \ \ \ \ \left[By Rationalization\right] \\ = \dfrac{1.6\left(\sqrt{3}+1\right)}{\left(\sqrt{3}\right)^2 - \left(1\right)^2} \\ \dfrac{1.6\left(\sqrt{3}+1\right)}{2} = 0.8\left(\sqrt{3}+1\right)\\$ Therefore, the height of the pedestal is $0.8\left(\sqrt{3}+1\right)m$

9   The angle of the elevation of the top of a building from the foot of the tower is $30^o$ and the angle of the elevation of the top of the tower from the foot of the building is $60^o$. If the tower is $50 m$ high,find the height of the building.

##### Solution : Let AB be the building CD be the tower. $In\Delta CDB, \\ \dfrac{CD}{BD} = \tan60^o \\ \dfrac{50}{BD}= \sqrt{3} \\ BD= \dfrac{50}{\sqrt{3}} \\ \ \ \ \ \ \\ In \Delta ABD, \\ \dfrac{AB}{BD} = \tan30^o \\ AB= \dfrac{50}{\sqrt{3}}*\dfrac{1}{\sqrt{3}} = \dfrac{50}{3} = 16\dfrac{2}{3} \\$ Threrefore, the height of the building is $16\dfrac{2}{3}m$.

10   Two poles of equal heights are standing opposite each other on either side of the road, which is $80 m$ wide. From a point between them on the road, the angles of elevation of the top of the poles are $60^o$ and $30^o$,respectively. Find the height of the poles and the distance of the point from the poles.

##### Solution : Let AB and CD be the poles and $O$ is the point from where the elevation angles are measured. $In \Delta CDO, \\ \dfrac{AB}{BO} = \tan60^o \\ \dfrac{AB}{BO} = \sqrt{3} \\ BO = \dfrac{AB}{\sqrt{3}} \\ \\ In \Delta CDO, \\ \dfrac{CD}{DO} = \tan30^o \\ \dfrac{CD}{80-BO} = \dfrac{1}{\sqrt{3}} \\ CD\sqrt{3} = 80-BO \\ Cd\sqrt{3} = 80-\dfrac{AB}{\sqrt{3}} \\ CD\sqrt{3}+\dfrac{AB}{\sqrt{3}} = 80 \\ Since the poles are of equal heights. \\ CD=AB \\ CD\left[\sqrt{3}+\dfrac{1}{\sqrt{3}}\right] = 80\\ CD\left(\dfrac{3+1}{\sqrt{3}}\right) = 80 \\ CD = 20\sqrt{3}m\\ BO = \dfrac{AB}{\sqrt{3}} = \dfrac{CD}{\sqrt{3}} = \left(\dfrac{20\sqrt{3}}{\sqrt{}3}\right)m = 20 m\\ DO = BD-BO = \left(80-20\right)m = 60 m$ Therefore,the height of the poles is $20\sqrt{3}$ and the point is $20 m$ and $60 m$ far from these poles.

11   A TV tower stands vertically on a bank of a canal.From a point on the other bank directly opposite the lower the angle of elevation of the top of the tower is $60^o$ . From another point $20 m$ away from this point on the line joining this point to the foot of the tower,the angle of elevation of the top of the tower is $30^o$ . Find the height of the tower and the width of the canel. ##### Solution :

$In \Delta ABC,\\ \dfrac{AB}{BC} = \tan60^o \\ \dfrac{AB}{BC} = \sqrt{3} \\ BC = \dfrac{BC}{\sqrt{}3} \\ \ \ \ \ \\ In \Delta ABD , \\ \dfrac{AB}{BD} = \tan30^o \\ \dfrac{Ab}{BC+CD} = \dfrac{1}{\sqrt{3}} \\ \dfrac{AB}{\dfrac{AB}{\sqrt{3}}+20} = \dfrac{1}{\sqrt{3}} \\ \dfrac{AB\sqrt{3}}{AB+20\sqrt{3}} = \dfrac{1}{\sqrt{3}} \\ 3AB = AB+20\sqrt{3} \\ 2AB= 20\sqrt{3} \\ AB = 10\sqrt{3}m \\ BC = \dfrac{AB}{\sqrt{3}} = \left(dfrac{10\sqrt{3}}{sqrt{3}}\right)m = 10 m$ Therefore, the height of the tower is $10\sqrt{3} m$ and the width of the canal is $10 m$ .

12   From the top of a $7 m$ high building , the angle of elevation of the top of a cable tower is $60^o$ and the angle of depression of its foot is $45^o$ . Determine the height of the tower.

##### Solution : 