Mathematics · Some Applications of Trigonometry · NCERT Exercises

Exercise 9.1

Complete, independently verified solutions for NCERT Exercise 9.1.

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Mathematics · Chapter 9 · NCERT Exercises

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Verified NCERT questions and solutionsSolutions are checked independently and follow the supplied NCERT chapter edition.

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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°.

sin⁡30∘=ABAC\sin30^\circ=\frac{AB}{AC}
12=AB20\frac12=\frac{AB}{20}
AB=10 mAB=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.

Broken treeThe remaining trunk is vertical, the broken section touches the ground eight metres from its foot and makes a 30 degree angle there.8 m30°ABC
Broken tree — explanatory sketch (not to scale).

Solution

Board-exam working:

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

tan⁡30∘=ABBC\tan30^\circ=\frac{AB}{BC}
13=AB8\frac1{\sqrt3}=\frac{AB}{8}
AB=83 mAB=\frac8{\sqrt3}\text{ m}
cos⁡30∘=BCAC\cos30^\circ=\frac{BC}{AC}
32=8AC\frac{\sqrt3}{2}=\frac8{AC}
AC=163 mAC=\frac{16}{\sqrt3}\text{ m}
Original height=AB+AC\text{Original height}=AB+AC
=83+163=243=83 m=\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.

Two playground slidesSeparate right triangles show a 1.5 metre rise at 30 degrees and a 3 metre rise at 60 degrees.1.5 mlength 130°3 mlength 260°
Two playground slides — explanatory sketch (not to scale).

Solution

Board-exam working:

(i) Slide for younger children:

sin⁡30∘=1.5L1\sin30^\circ=\frac{1.5}{L_1}
12=1.5L1\frac12=\frac{1.5}{L_1}
L1=3 mL_1=3\text{ m}

(ii) Slide for older children:

sin⁡60∘=3L2\sin60^\circ=\frac3{L_2}
32=3L2\frac{\sqrt3}{2}=\frac3{L_2}
L2=63=23 mL_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.

Tower seen from 30 metresA right triangle joins the top of a vertical tower to an observation point 30 metres from its base; the ground angle is 30 degrees.h30 m30°
Tower seen from 30 metres — explanatory sketch (not to scale).

Solution

Board-exam working:

Let h be the height of the tower.

tan⁡30∘=h30\tan30^\circ=\frac{h}{30}
13=h30\frac1{\sqrt3}=\frac{h}{30}
h=303=103 mh=\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.

Kite and taut stringThe kite is 60 metres above the ground and its taut string makes a 60 degree angle with the ground.60 mstring length60°
Kite and taut string — explanatory sketch (not to scale).

Solution

Board-exam working:

Let L be the string length.

sin⁡60∘=60L\sin60^\circ=\frac{60}{L}
32=60L\frac{\sqrt3}{2}=\frac{60}{L}
L=1203=403 mL=\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.

Boy walking towards a buildingA thirty metre building is viewed from two eye-level positions 1.5 metres above ground, with angles of elevation 30 and 60 degrees.30 m1.5 m30°60°PSA
Boy walking towards a building — explanatory sketch (not to scale).

Solution

Board-exam working:

Height of the building above the boy's eyes:

30−1.5=28.5 m30-1.5=28.5\text{ m}

Let the initial distance be x m.

tan⁡30∘=28.5x\tan30^\circ=\frac{28.5}{x}
13=28.5x\frac1{\sqrt3}=\frac{28.5}{x}
x=28.53 mx=28.5\sqrt3\text{ m}

Let the final distance be y m.

tan⁡60∘=28.5y\tan60^\circ=\frac{28.5}{y}
3=28.5y\sqrt3=\frac{28.5}{y}
y=28.53=9.53 my=\frac{28.5}{\sqrt3}=9.5\sqrt3\text{ m}
Distance walked=x−y\text{Distance walked}=x-y
=28.53−9.53=193 m=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.

Transmission tower on a buildingThe bottom and top of a transmission tower on a 20 metre building are seen from one ground point at 45 and 60 degrees.20 mh45°60°
Transmission tower on a building — explanatory sketch (not to scale).

Solution

Board-exam working:

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

Using the bottom of the tower:

tan⁡45∘=20x\tan45^\circ=\frac{20}{x}
x=20 mx=20\text{ m}

Using the top of the tower:

tan⁡60∘=20+h20\tan60^\circ=\frac{20+h}{20}
3=20+h20\sqrt3=\frac{20+h}{20}
20+h=20320+h=20\sqrt3
h=20(3−1) mh=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.

Statue on a pedestalA 1.6 metre statue stands on a pedestal of height h, with 45 and 60 degree sight lines from the same ground point.h1.6 m45°60°
Statue on a pedestal — explanatory sketch (not to scale).

Solution

Board-exam working:

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

tan⁡45∘=hx\tan45^\circ=\frac{h}{x}
x=hx=h

Using the top of the statue:

tan⁡60∘=h+1.6x\tan60^\circ=\frac{h+1.6}{x}
3=h+1.6h\sqrt3=\frac{h+1.6}{h}
h(3−1)=1.6h(\sqrt3-1)=1.6
h=1.63−1h=\frac{1.6}{\sqrt3-1}
h=1.6(3+1)2=0.8(3+1) mh=\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.

Building and tower viewed from each other's feetA shorter building and a 50 metre tower stand apart, with lines of sight at 30 and 60 degrees from their opposite feet.xh50 m30°60°
Building and tower viewed from each other's feet — explanatory sketch (not to scale).

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:

tan⁡60∘=50x\tan60^\circ=\frac{50}{x}
3=50x\sqrt3=\frac{50}{x}
x=503 mx=\frac{50}{\sqrt3}\text{ m}

Using the building:

tan⁡30∘=hx\tan30^\circ=\frac{h}{x}
13=h50/3\frac1{\sqrt3}=\frac{h}{50/\sqrt3}
h=503 mh=\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.

Equal poles across an 80 metre roadEqual-height poles stand at opposite edges of an 80 metre road; a point between them sees their tops at 60 and 30 degrees.80 mhh60°30°P
Equal poles across an 80 metre road — explanatory sketch (not to scale).

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.

tan⁡60∘=hx\tan60^\circ=\frac{h}{x}
h=x3(1)h=x\sqrt3\qquad\text{(1)}
tan⁡30∘=h80−x\tan30^\circ=\frac{h}{80-x}
13=h80−x\frac1{\sqrt3}=\frac{h}{80-x}
80−x=h3(2)80-x=h\sqrt3\qquad\text{(2)}

Substitute h = x√3 in (2):

80−x=3x80-x=3x
x=20 mx=20\text{ m}
80−x=60 m80-x=60\text{ m}
h=203 mh=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 across a canalA vertical tower AB is viewed from point C on the opposite bank at 60 degrees and from point D, twenty metres farther away, at 30 degrees; the full D-C-B baseline is visible.h20 m30°60°ABCD
TV tower across a canal — complete redraw of the supplied NCERT figure (not to scale).

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:

tan⁡60∘=hx\tan60^\circ=\frac{h}{x}
h=x3(1)h=x\sqrt3\qquad\text{(1)}

From point D:

tan⁡30∘=hx+20\tan30^\circ=\frac{h}{x+20}
13=hx+20\frac1{\sqrt3}=\frac{h}{x+20}
x+20=h3(2)x+20=h\sqrt3\qquad\text{(2)}

Substitute (1) in (2):

x+20=3xx+20=3x
x=10 mx=10\text{ m}
h=103 mh=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.

Cable tower viewed from a buildingFrom the top of a 7 metre building, one sight line rises 60 degrees to a taller cable tower and another falls 45 degrees to its foot.7 mH60°45°
Cable tower viewed from a building — explanatory sketch (not to scale).

Solution

Board-exam working:

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

tan⁡45∘=7x\tan45^\circ=\frac7x
x=7 mx=7\text{ m}

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

tan⁡60∘=h7\tan60^\circ=\frac h7
3=h7\sqrt3=\frac h7
h=73 mh=7\sqrt3\text{ m}
Tower height=7+h\text{Tower height}=7+h
=7+73=7(1+3) m=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.

Two ships viewed from a lighthouseA 75 metre lighthouse sees a nearer ship at 45 degrees depression and a farther ship on the same side at 30 degrees depression.75 m30°45°Far shipNear ship
Two ships viewed from a lighthouse — explanatory sketch (not to scale).

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:

tan⁡45∘=75x\tan45^\circ=\frac{75}{x}
x=75 mx=75\text{ m}

For the farther ship:

tan⁡30∘=75y\tan30^\circ=\frac{75}{y}
13=75y\frac1{\sqrt3}=\frac{75}{y}
y=753 my=75\sqrt3\text{ m}
Distance between ships=y−x\text{Distance between ships}=y-x
=753−75=75(3−1) m=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.

Balloon seen at two anglesThe girl's eye is 1.2 metres above ground. A balloon at constant height 88.2 metres moves from the position seen at 60 degrees to the position seen at 30 degrees; all labels fit inside the sketch.1.2 m87 m88.2 mdistance travelled60°30°
Balloon seen at two angles — complete redraw of the supplied NCERT figure (not to scale).

Solution

Board-exam working:

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

88.2−1.2=87 m88.2-1.2=87\text{ m}

Let the initial horizontal distance be x m.

tan⁡60∘=87x\tan60^\circ=\frac{87}{x}
x=873=293 mx=\frac{87}{\sqrt3}=29\sqrt3\text{ m}

Let the later horizontal distance be y m.

tan⁡30∘=87y\tan30^\circ=\frac{87}{y}
y=873 my=87\sqrt3\text{ m}
Distance travelled=y−x\text{Distance travelled}=y-x
=873−293=583 m=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.

Car approaching a towerA car is observed from the top of a tower at 30 degrees depression and six seconds later at 60 degrees depression before reaching its foot.h30°60°First position6 s later
Car approaching a tower — explanatory sketch (not to scale).

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:

tan⁡30∘=hx\tan30^\circ=\frac hx
x=h3x=h\sqrt3

At the second position:

tan⁡60∘=hy\tan60^\circ=\frac hy
y=h3y=\frac h{\sqrt3}

Distance travelled in 6 seconds:

x−y=h3−h3=2h3x-y=h\sqrt3-\frac h{\sqrt3}=\frac{2h}{\sqrt3}

The remaining distance is:

y=h3y=\frac h{\sqrt3}

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

Required time=62=3 seconds\text{Required time}=\frac62=3\text{ seconds}

Final answer

The car will reach the tower in 3 seconds.

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