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 :

Let $AB$ be a building and Cd be a cable tower. $ In \Delta ABD, \\ \dfrac{AB}{BD} = \tan45^o \\ \dfrac{7}{BD} = 1 \implies BD = 7m\\ \\ In \Delta ACE, \\ AE = BD = 7m \\ \\ \dfrac{CE}{AE} = \tan60^o \\ \dfrac{CE}{7 } = \sqrt{3} \\ CE = 7\sqrt{3}m \\ CD = CE +ED = \left(7\sqrt{3}+7\right) m \\ \ \ \ \ \ \ \ = 7\left(\sqrt{3}+1\right) m \\ $ Therefore, the height of the cable tower is $\left(7\sqrt{3}+7\right) m.

13   As observed from the top of a $75 m $ high lighthouse from the sea-level, the angles of depression of two ships are $30^o$ and $45^o$ . If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.

Solution :

Let $AB$ be the lighthouse and the two ships be at point C and D respectively. $ In \Delta ABC, \\ \dfrac{AB}{BC} = \tan45^o \\ \dfrac{75}{BC} = 1 \\ BC = 75 m \\ \\ In \Delta ABD, \\ \dfrac{AB}{BD} = \tan30^o \\ \dfrac{75}{BC+CD} = \dfrac{1}{\sqrt{3}} \\ \dfrac{75}{75+CD} = \dfrac{1}{\sqrt{3}} \\ 75\sqrt{3} = 75+CD \\ 75\left(\sqrt{3}-1\right)m = CD \\$ Therefore, the distance between the two ships is $75\left(\sqrt{3}=1\right) m $.

14   A $1.2 m$ tall girl spots a balloon moving with the wind in a horizontal line at a height of $88.2 m$ from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is $60^o$ . After some time, the angle of elevation reduces to $30^o$. Find the distance travelled by the balloon during the interval.

Solution :

Let the initial position of $A$ of balloon change to $B$ after some time $CD$ be the girl. $ In \Delta ACE, \\ \dfrac{AE}{CE} = \tan60^o \\ \dfrac{88.2-1.2}{CE} = \sqrt{3} \\ \dfrac{87}{CE} = \sqrt{3} \\ CE = \dfrac{87}{\sqrt{3}} = 29\sqrt{3}m \\ \ \ \ \ \ \\ In \Delta BCG, \\ \dfrac{BG}{CG} = \tan30^o \\ \dfrac{BH-GH}{CG} = \dfrac{1}{\sqrt{3}} \\ \dfrac{88.2 - 1.2}{CG} = \dfrac{1}{\sqrt{3}} \\ 87\sqrt{3}m = CG\\ Distance travelled by balloon = EG = CG- CE \\ = \left(87\sqrt{3} - 29\sqrt{3}\right ) M \\ = 58\sqrt{3}m. $

15   A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car as an angle of depression of 30°, which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be 60°. Find the time taken by the car to reach the foot of the tower from this point.

Solution :

Let AB be the tower. Initial position of the car is C, which changes to D after six seconds. $In \Delta ADB , \\ \dfrac{AB}{DB} = \tan60^o \\ \dfrac{AB}{DB} = 3 \\ DB = \dfrac{AB}{\sqrt{3}} \\ \\ In \Delta ABC, \\ \dfrac{AB}{BC} = \tan30^o \\ dfrac{AB}{BD+DC} = \dfrac{1}{\sqrt{3}} \\ AB 3 = BD + DC \\ AB\sqrt{3} = \dfrac{AB}{\sqrt{3}} + DC \\ DC= AB\sqrt{3}-\dfrac{AB}{\sqrt{3}}= Ab\left(\sqrt{3}-\dfrac{1}{\sqrt{3}}\right) \\ = \dfrac{2AB}{\sqrt{3}} \\$ Time taken by the car to travel distance DC $\left(i.e. \dfrac{2AB}{\sqrt{3}}\right) = 6 seconds $ \\ Time taken by the car to travel distance DB $\left(i.e.,\dfrac{AB}{\sqrt{3}}\right) = \dfrac{6}{\dfrac{2AB}{\sqrt{3}}}*\dfrac{AB}{\sqrt{3}} = \dfrac{6}{2} = 3 seconds. $.

16   The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is 6 m.

Solution :

17   The angles of elevation of the top of a tower from two points at a distance of $4 m$ and $9 m$ from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is $6 m$.

Solution :

$In \Delta AQS, $

the other will be

$\left(\cot \theta \right) \\ \dfrac{AQ^2}{36} = 1 $

However, height cannot be negative.

Let $A Q$ be the tower and R, S are the points $4 m$, $9 m$ away from the base of the tower respectively.

$\left(\dfrac{AQ}{4}\right)$

$\tan \left(90 - \theta \right) $

On multiplying equations

$\dfrac{A Q}{QR} $

$\tan \theta $

Therefore, if one angle is $\theta$ ,

$\left(\tan\theta \right) $

$\cot \theta $

$AQ = \sqrt{36} =+6$

we obtain

$\dfrac{AQ}{SQ} =$

$ In \Delta AQR, $

$\dfrac{AQ}{4} =$

Therefore, the height of the tower is $6 m $.

The angles are complementary.

$\left(\dfrac{AQ}{9}\right) =$

AQ^2 =36

$ \dfrac{AQ}{9} =$

$\left(i \right) and \left(ii \right) $

$ = \tan \theta$