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Learn Exercise 9.1 with Free Lessons & Tips

Question 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° (see the figure).

Solution 1:

It can be observed from the figure that AB is the pole.

In ΔABC,

$\frac{AB}{AC}=sin30^{\circ}$

$\frac{AB}{20}=\frac{1}{2}$

$AB=\frac{20}{2}=10$

Therefore, the height of the pole is 10 m.

Comments

Tasneem

Question 2:

A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle $30^{\circ}$ 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 2:

Let AC was the original tree. Due to storm, it was broken into two parts. The broken part is making 30° with the ground.

In ,

$\frac{BC}{A'C}=tan30^{\circ}$

$\frac{BC}{8}=\frac{1}{\sqrt{}3}$

$BC=\frac{8}{\sqrt{3}}m$

$\frac{A'B}{A'C}=cos30^{\circ}$

$\frac{8}{A'B}=\frac{\sqrt{}3}{2}$

$A'B=\frac{16}{\sqrt{3}}m$

Height of tree = + BC

$\left ( \frac{16}{\sqrt{3}}+\frac{8}{\sqrt{3}} \right )$ $m=\frac{24}{\sqrt{3}}m$ $=8\sqrt{3}$

Hence, the height of the tree is.

Comments

Tasneem

Question 3:

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

Solution 3:

It can be observed that AC and PR are the slides for younger and elder children respectively.

In ΔABC,

$\frac{AB}{AC}=sin30^{\circ}$

$\frac{1.5}{AC}=\frac{1}{2}$

AC=3m

In ΔPQR,

$\frac{PQ}{PR}=sin60^{\circ}$

$\frac{3}{PR}=\frac{\sqrt{3}}{2}$

$PR=\frac{6}{\sqrt{3}}=2\sqrt{3}m$

Therefore, the lengths of these slides are 3 m and .

Comments

Tasneem

Question 4:

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

Solution 4:

Let AB be the tower and the angle of elevation from point C (on ground) is

30°.

In ΔABC,

$\frac{AB}{BC}=tan30^{\circ}$

$\frac{AB}{30}=\frac{1}{\sqrt{3}}$ , $\therefore AB=\frac{30}{\sqrt{3}}=10\sqrt{3}m$

Therefore, the height of the tower is.

Comments

Tasneem

Question 5:

A kite is flying at a height of 60 m above 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 $60^{\circ}$ . Find the length of the string, assuming that there is no slack in the string.

Solution 5:

Let K be the kite and the string is tied to point P on the ground.

In ΔKLP,

$\frac{KL}{KP}=sin60^{\circ}$

$\frac{60}{KP}=\frac{\sqrt{3}}{2}$

$KP=\frac{120}{\sqrt{3}}=40\sqrt{3}m$

Hence, the length of the string is.

Comments

Tasneem

Question 6:

A 1.5 m 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 $30^{\circ}$ to $60^{\circ}$ as he walks towards the building. Find the distance he walked towards the building.

Solution 6:

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

= (30 − 1.5) m = 28.5 m =

In ΔPAR,

$\frac{PR}{AR}=tan30^{\circ}$

$\frac{57}{2AR}=\frac{1}{\sqrt{3}}, AR=\left ( \frac{57}{2}\sqrt{3} \right )m$

In ΔPRB,

$\frac{PR}{QR}=tan60^{\circ}$

$\frac{57}{2BR}=\sqrt{3}$

$BR=\frac{57}{2\sqrt{3}}=\left ( 19\frac{\sqrt{3}}2{} \right )m$

ST = AB

$AR-BR=\left ( \frac{57\sqrt{3}}2{-\frac{19\sqrt{3}}{2}} \right )m$

$=\left ( \frac{38\sqrt{3}}{2} \right )m=19\sqrt{3}m$

Hence, he walked towards the building.

Comments

Tasneem

Question 7:

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

Solution 7:

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 ΔBCD,

$\frac{AC}{CD}=tan60^{\circ}$

$\frac{AB+BC}{CD}=\sqrt{3}$

$\frac{AB+20}{20}=\sqrt{3}$

$AB=\left ( 20\sqrt{3}-20 \right )m$

$=20\left ( \sqrt{3}-1 \right )m$

Therefore, the height of the transmission tower is m.

Comments

Tasneem

Question 8:

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 $45^{\circ}$ . Find the height of the pedestal.

Solution 8:

Let AB be the statue, BC be the pedestal, and D be the point on the ground from where the elevation angles are to be measured.

In ΔBCD,

$\frac{BC}{CD}=tan 45^{\circ}$

$\frac{BC}{CD}=1$

BC=CD

In ΔACD,

$\frac{AB+BC}{CD}=tan60^{\circ}$

$g_white \frac{AB+BC}{ BC}=\sqrt{3}$

$1.6+BC=BC\sqrt{3}$

$BC\left ( \sqrt{3}-1\right )=1.6$

$BC=\frac{1.6\left ( \sqrt{3}+1 \right )}{\left ( \sqrt{3}-1 \right )\left ( \sqrt{3}+1 \right )}$

$=\frac{1.6(\sqrt{3}+1)} { \left ( \sqrt{3} \right )^{2}-\left ( 1 \right ) ^{2} }$ $=\frac{1.6(\sqrt{3}+1)} {2 }=0.8(\sqrt{3}+1)$

Therefore, the height of the pedestal is 0.8m.

Comments

Tasneem

Question 9:

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

Solution 9:

Let AB be the building and CD be the tower.

In ΔCDB,

$\frac{CD}{BD}=tan60^{\circ}$

$\frac{50}{BD}=\sqrt{3}$

$\therefore BD=\frac{50}{\sqrt{3}}$

In ΔABD,

$\frac{AB}{BD}=tan30^{\circ}$

$AB=\frac{50}{\sqrt{3}}\times \frac{1}{\sqrt{3}}$ $=\frac{50}{3}=16\frac{2}{3}$

Therefore, the height of the building is.

Comments

Tasneem

Question 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 $60^{\circ}$ and $30^{\circ}$ , respectively. Find the height of the poles and the distances of the point from the poles.

Solution 10:

Let AB and CD be the poles and O is the point from where the elevation angles are measured.

In ΔABO, $\frac{AB}{BO}=tan60^{\circ}$

$\frac{AB}{BO}=\sqrt{3}$

In ΔCDO,

$\frac{CD}{DO}=tan30^{\circ}$

$\frac{CD}{80-BO}=\frac{1}{\sqrt{3}}$

$CD\sqrt{3}=80-BO$

$CD\sqrt{3}=80-\frac{AB}{\sqrt{3}}$

$CD\sqrt{3}+\frac{AB}{\sqrt{3}}=80$

Since the poles are of equal heights,

CD = AB

$CD\left \lfloor \sqrt{3}+\frac{1}{\sqrt{3}} \right \rfloor=80$

$CD=\left ( 3+\frac{1}{\sqrt{3}} \right )=80$

$CD=2\sqrt{3}m$

$BO=\frac{AB}{\sqrt{3}}=\frac{CD}{\sqrt{3}}=\left ( \frac{20\sqrt{3}}{\sqrt{3}} \right )m$

=20 m

DO = BD − BO = (80 − 20) m = 60 m

Therefore, the height of poles isand the point is 20 m and 60 m far from these poles.

Comments

Tasneem

Question 11:

A TV tower stands vertically on a bank of a canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is $60^{\circ}$ . From another point 20 m away from this point on the line joing this point to the foot of the tower, the angle of elevation of the top of the tower is $30^{\circ}$ (see the figure). Find the height of the tower and the width of the canal.

Solution 11:

In ΔABC,

$\frac{AB}{BC}=tan60^{\circ}$

$\frac{AB}{BC}=\sqrt{3}$

$BC=\frac{AB}{\sqrt{3}}$

In ΔABD,

$\frac{AB}{BD}=tan30^{\circ}$

$\frac{AB}{BC+CD}=\frac{1}{\sqrt{3}}$

$\frac{AB}{\frac{AB}{\sqrt{3}}+20}=\frac{1}{\sqrt{3}}$

$\frac{AB\sqrt{3}}{AB+20\sqrt{3}}=\frac{1}{\sqrt{3}}$

$3AB=AB+20\sqrt{3}$

$2AB=20\sqrt{3}$

$AB=10\sqrt{3}m$

$BC=\frac{AB}{\sqrt{3}}=\left ( \frac{10\sqrt{3}}{\sqrt{3}} \right )m$ =10m

Therefore, the height of the tower is m and the width of the canal is 10 m.

Comments

Tasneem

Question 12:

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

Solution 12:

Let AB be a building and CD be a cable tower.

In ΔABD,

$\frac{AB}{BD}=tan45^{\circ}$

$\frac{7}{BD}=1$ , BD= 7m

In ΔACE,

AE = BD = 7 m

$\frac{CE}{AE}=tan60^{\circ}$

$\frac{CE}7{}=\sqrt{3}$ , $\therefore CE=7\sqrt{3}m$

$CD= CE+ED=\left ( 7\sqrt{3}+7 \right )m =7\left ( \sqrt{3}+1 \right )m$

Therefore, the height of the cable tower is.

Comments

Tasneem

Question 13:

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

Solution 13:

Let AB be the lighthouse and the two ships be at point C and D respectively.

In ΔABC,

$\frac{AB}{BC}=tan45^{\circ}$

$\frac{75}{BC}=1, BC=75m$

In ΔABD,

$\frac{AB}{BD}=tan30^{\circ}$

$\frac{75}{BC+CD}=\frac{1}{\sqrt{3}}$

$\frac{75}{75+CD}=\frac{1}{\sqrt{3}}$

$75\sqrt{3}=75+CD$

$75\left ( \sqrt{3}-1 \right )m=CD$

Therefore, the distance between the two ships ism.

Comments

Tasneem

Question 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 $60^{\circ}$ . After some time, the angle of elevation reduces to $30^{\circ}$ (see the figure). Find the distance travelled by the balloon during the interval.

Solution 14:

Let the initial position A of balloon change to B after some time and CD be the girl.

In ΔACE,

$\frac{AE}{CE}=tan60^{\circ}$

$\frac{AF-EF}{CE}=tan60^{\circ}$

$\frac{88.2-1.2}{CE}=\sqrt{3}$ $\frac{87}{CE}=\sqrt{3}$

$CE=\frac{87}{\sqrt{3}}=29\sqrt{3}m$

In ΔBCG,

$\frac{BG}{CG}=tan30^{\circ}$

$\frac{88.2-1.2}{CG}=\frac{1}{\sqrt{3}}$

$87\sqrt{3}m=CG$

Distance travelled by balloon = EG = CG − CE

Comments

Tasneem

Question 15:

A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of $30^{\circ}$ , 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 $60^{\circ}$ . Find the time taken by the car to reach the foot of the tower from this point.

Solution 15:

Let AB be the tower.

Initial position of the car is C, which changes to D after six seconds.

In ΔADB,

$\frac{AB}{DB}=tan60^{\circ}$

$\frac{AB}{BD}=\sqrt{3}$

$DB=\frac{AB}{\sqrt{3}}$

In ΔABC,

$\frac{AB}{BC}=tan30^{\circ}$

$\frac{AB}{BD+DC}=\frac{1}{\sqrt{3}}$

$AB\sqrt{3}=BD+DC$

$AB\sqrt{3}=\frac{AB}{\sqrt{3}}+DC$

$DC=AB-\frac{AB}{\sqrt{3}}=AB\left ( \sqrt{3}-\frac{1}{\sqrt{3}} \right )$

$=\frac{2AB}{\sqrt{3}}$

Time taken by the car to travel distance DC= 6 seconds

Time taken by the car to travel distance DB

Comments

Tasneem

Question 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 16:

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

The angles are complementary. Therefore, if one angle is θ, the other will be 90 − θ.

In ΔAQR,

$\frac{AQ}{QR}=tan\theta$

$\frac{AQ}{4}=tan\theta..........(i)$

In ΔAQS,

$\frac{AQ}{SQ}=tan(90-\theta )$

$\frac{AQ}{9}=cot\theta$ $g_white ......(ii)$

On multiplying equations (i) and (ii), we get

$\left ( \frac{AQ}{4} \right )\left ( \frac{AQ}{9} \right )=(tan\theta )(cot\theta )$

$\frac{AQ^{2}}{36}=1, AQ=\sqrt{}36\therefore AQ=\pm 6$

However, height cannot be negative.

Therefore, the height of the tower is 6 m.

Comments

Tasneem

All Exercises - Chapter 9 - Some Applications of Trigonometry

Exercise 9.1

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Learn Exercise 9.1 with Free Lessons & Tips

All Exercises - Chapter 9 - Some Applications of Trigonometry

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