NCERT Exercise 9.1 — Some Applications of Trigonometry

Question

Question 1

circus rope · pole height

A circus artist is climbing a 20 m long rope, tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole if the rope makes an angle of 30° with the ground.

A 20 metre rope AC tied from the top of pole AB to the ground at C, making a 30 degree angle. — Figure 9-11 from the supplied NCERT chapter.

Solution

Board-exam working:

In right triangle ABC, AC = 20 m and ∠ACB = 30°.

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

\frac12=\frac{AB}{20}

AB=10\text{ m}

Final answer

The pole is 10 m high.

Question

Question 2

broken tree · right triangle

A tree breaks due to a storm and the broken part bends so that its top touches the ground, making an angle of 30° with it. The distance from the foot of the tree to the point where the top touches the ground is 8 m. Find the original height of the tree.

Solution

Board-exam working:

Let AB be the remaining upright part, AC the broken part and BC = 8 m.

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

\frac1{\sqrt3}=\frac{AB}{8}

AB=\frac8{\sqrt3}\text{ m}

\cos30^\circ=\frac{BC}{AC}

\frac{\sqrt3}{2}=\frac8{AC}

AC=\frac{16}{\sqrt3}\text{ m}

\text{Original height}=AB+AC

=\frac8{\sqrt3}+\frac{16}{\sqrt3}=\frac{24}{\sqrt3}=8\sqrt3\text{ m}

Final answer

The original height of the tree was 8√3 m.

Question

Question 3

slides · length

A contractor plans two slides. One has a height of 1.5 m and is inclined at 30° to the ground; the other has a height of 3 m and is inclined at 60°. Find the length of each slide.

Solution

Board-exam working:

(i) Slide for younger children:

\sin30^\circ=\frac{1.5}{L_1}

\frac12=\frac{1.5}{L_1}

L_1=3\text{ m}

(ii) Slide for older children:

\sin60^\circ=\frac3{L_2}

\frac{\sqrt3}{2}=\frac3{L_2}

L_2=\frac6{\sqrt3}=2\sqrt3\text{ m}

Final answer

The slide lengths are 3 m and 2√3 m.

Question

Question 4

tower · angle of elevation

The angle of elevation of the top of a tower from a point 30 m away from its foot is 30°. Find the height of the tower.

Solution

Board-exam working:

Let h be the height of the tower.

\tan30^\circ=\frac{h}{30}

\frac1{\sqrt3}=\frac{h}{30}

h=\frac{30}{\sqrt3}=10\sqrt3\text{ m}

Final answer

The tower is 10√3 m high.

Question

Question 5

kite · string length

A kite is flying at a height of 60 m above the ground. Its string is inclined at 60° to the ground. Find the length of the string, assuming there is no slack.

Solution

Board-exam working:

Let L be the string length.

\sin60^\circ=\frac{60}{L}

\frac{\sqrt3}{2}=\frac{60}{L}

L=\frac{120}{\sqrt3}=40\sqrt3\text{ m}

Final answer

The string is 40√3 m long.

Question

Question 6

walking distance · two observation points

A 1.5 m tall boy stands at some distance from a 30 m building. The angle of elevation from his eyes to the top increases from 30° to 60° as he walks towards the building. Find the distance he walked.

Solution

Board-exam working:

Height of the building above the boy's eyes:

30-1.5=28.5\text{ m}

Let the initial distance be x m.

\tan30^\circ=\frac{28.5}{x}

\frac1{\sqrt3}=\frac{28.5}{x}

x=28.5\sqrt3\text{ m}

Let the final distance be y m.

\tan60^\circ=\frac{28.5}{y}

\sqrt3=\frac{28.5}{y}

y=\frac{28.5}{\sqrt3}=9.5\sqrt3\text{ m}

\text{Distance walked}=x-y

=28.5\sqrt3-9.5\sqrt3=19\sqrt3\text{ m}

Final answer

The boy walked 19√3 m.

Question

Question 7

transmission tower · building

From a point on the ground, the angles of elevation of the bottom and top of a transmission tower fixed on a 20 m high building are 45° and 60°, respectively. Find the height of the tower.

Solution

Board-exam working:

Let x be the horizontal distance and h be the transmission tower height.

Using the bottom of the tower:

\tan45^\circ=\frac{20}{x}

x=20\text{ m}

Using the top of the tower:

\tan60^\circ=\frac{20+h}{20}

\sqrt3=\frac{20+h}{20}

20+h=20\sqrt3

h=20(\sqrt3-1)\text{ m}

Final answer

The transmission tower is 20(√3 − 1) m high.

Question

Question 8

statue · pedestal

A statue 1.6 m tall stands on a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and that of the top of the pedestal is 45°. Find the height of the pedestal.

Solution

Board-exam working:

Let the pedestal height be h m and the horizontal distance be x m.

\tan45^\circ=\frac{h}{x}

x=h

Using the top of the statue:

\tan60^\circ=\frac{h+1.6}{x}

\sqrt3=\frac{h+1.6}{h}

h(\sqrt3-1)=1.6

h=\frac{1.6}{\sqrt3-1}

h=\frac{1.6(\sqrt3+1)}2=0.8(\sqrt3+1)\text{ m}

Final answer

The pedestal is 0.8(√3 + 1) m high.

Question

Question 9

building and tower · reciprocal angles

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

Solution

Board-exam working:

Let the distance between the building and tower be x m and the building height be h m.

Using the 50 m tower:

\tan60^\circ=\frac{50}{x}

\sqrt3=\frac{50}{x}

x=\frac{50}{\sqrt3}\text{ m}

Using the building:

\tan30^\circ=\frac{h}{x}

\frac1{\sqrt3}=\frac{h}{50/\sqrt3}

h=\frac{50}{3}\text{ m}

Final answer

The building is 50/3 m high.

Question

Question 10

equal poles · road width

Two poles of equal heights stand opposite each other on either side of an 80 m wide road. From a point between them, the angles of elevation of their tops are 60° and 30°. Find the height of the poles and the distances of the point from them.

Solution

Board-exam working:

Let the pole height be h m and the distance from the pole seen at 60° be x m. The other distance is 80 − x.

\tan60^\circ=\frac{h}{x}

h=x\sqrt3\qquad\text{(1)}

\tan30^\circ=\frac{h}{80-x}

\frac1{\sqrt3}=\frac{h}{80-x}

80-x=h\sqrt3\qquad\text{(2)}

Substitute h = x√3 in (2):

80-x=3x

x=20\text{ m}

80-x=60\text{ m}

h=20\sqrt3\text{ m}

Final answer

Each pole is 20√3 m high; the point is 20 m and 60 m from the poles.

Question

Question 11

canal · two observation points

A TV tower stands vertically on a bank of a canal. From a point on the opposite bank directly opposite the tower, the angle of elevation is 60°. From another point 20 m farther away on the same line, the angle is 30°. Find the height of the tower and the width of the canal.

TV tower AB across a canal, viewed from C and from point D twenty metres farther away. — Figure 9-12 from the supplied NCERT chapter.

Solution

Board-exam working:

Let the canal width BC = x m and the tower height AB = h m. Then DB = x + 20 m.

From point C:

\tan60^\circ=\frac{h}{x}

h=x\sqrt3\qquad\text{(1)}

From point D:

\tan30^\circ=\frac{h}{x+20}

\frac1{\sqrt3}=\frac{h}{x+20}

x+20=h\sqrt3\qquad\text{(2)}

Substitute (1) in (2):

x+20=3x

x=10\text{ m}

h=10\sqrt3\text{ m}

Final answer

The canal is 10 m wide and the tower is 10√3 m high.

Question

Question 12

cable tower · elevation and depression

From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Find the height of the tower.

Solution

Board-exam working:

Let x be the horizontal distance. The observer is 7 m above ground.

\tan45^\circ=\frac7x

x=7\text{ m}

Let h be the part of the tower above the observer's level.

\tan60^\circ=\frac h7

\sqrt3=\frac h7

h=7\sqrt3\text{ m}

\text{Tower height}=7+h

=7+7\sqrt3=7(1+\sqrt3)\text{ m}

Final answer

The cable tower is 7(1 + √3) m high.

Question

Question 13

lighthouse · ships

From the top of a 75 m lighthouse, the angles of depression of two ships on the same side are 30° and 45°. One ship is exactly behind the other. Find the distance between the ships.

Solution

Board-exam working:

Let the nearer ship be x m from the lighthouse and the farther ship be y m away.

For the nearer ship:

\tan45^\circ=\frac{75}{x}

x=75\text{ m}

For the farther ship:

\tan30^\circ=\frac{75}{y}

\frac1{\sqrt3}=\frac{75}{y}

y=75\sqrt3\text{ m}

\text{Distance between ships}=y-x

=75\sqrt3-75=75(\sqrt3-1)\text{ m}

Final answer

The ships are 75(√3 − 1) m apart.

Question

Question 14

moving balloon · two angles of elevation

A 1.2 m tall girl spots a balloon moving horizontally at a height of 88.2 m. The angle of elevation from her eyes is initially 60° and later becomes 30°. Find the distance travelled by the balloon.

A balloon at height 88.2 metres viewed by a 1.2 metre tall girl at angles of 60 and 30 degrees. — Figure 9-13 from the supplied NCERT chapter.

Solution

Board-exam working:

Vertical height of the balloon above the girl's eyes:

88.2-1.2=87\text{ m}

Let the initial horizontal distance be x m.

\tan60^\circ=\frac{87}{x}

x=\frac{87}{\sqrt3}=29\sqrt3\text{ m}

Let the later horizontal distance be y m.

\tan30^\circ=\frac{87}{y}

y=87\sqrt3\text{ m}

\text{Distance travelled}=y-x

=87\sqrt3-29\sqrt3=58\sqrt3\text{ m}

Final answer

The balloon travelled 58√3 m.

Question

Question 15

moving car · angle of depression

A straight highway leads to the foot of a tower. A man at the top observes a car approaching at an angle of depression of 30°. Six seconds later, the angle of depression is 60°. Find the time the car will take to reach the tower from this second point, assuming uniform speed.

Solution

Board-exam working:

Let the tower height be h m. Let the car's initial and later distances from the tower be x m and y m.

At the first position:

\tan30^\circ=\frac hx

x=h\sqrt3

At the second position:

\tan60^\circ=\frac hy

y=\frac h{\sqrt3}

Distance travelled in 6 seconds:

x-y=h\sqrt3-\frac h{\sqrt3}=\frac{2h}{\sqrt3}

The remaining distance is:

y=\frac h{\sqrt3}

This is half the distance travelled in 6 seconds. At uniform speed, the time is also half.

\text{Required time}=\frac62=3\text{ seconds}

Final answer

The car will reach the tower in 3 seconds.