Class 10 Maths Chapter 9 Some Applications of Trigonometry NCERT Solutions in English

Contents hide
1 Height of the pole (circus rope)
2 Height of the broken tree
3 Length of the two slides
4 Height of the tower, 30 m away
5 Length of the kite string
6 Distance the boy walked
7 Height of the transmission tower
8 Height of the pedestal
9 Height of the building from the tower
10 Two equal poles across an 80 m road
11 TV tower and width of the canal
12 Height of the cable tower
13 Distance between two ships
14 Distance travelled by the balloon
15 Time for the car to reach the tower
16 In-text questions

Complete solutions to all 15 questions of Exercise 9.1 of the NCERT textbook Mathematics, Class X, Chapter 9, “Some Applications of Trigonometry”. Each question is given as printed in the current NCERT book (the misprint “joing” in Question 11 is corrected to “joining”), followed by a figure drawn to scale and a full step-by-step solution. The in-text questions of the chapter are answered at the end. Draw the figure yourself first: a labelled figure helps you set up each right triangle.

1 Height of the pole (circus rope)

Question 1

Q. 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 Fig. 9.11).

30° A B C rope AC = 20 m pole AB = ?
Figure, Question 1 (NCERT Fig. 9.11): rope AC = 20 m from the top A of pole AB to the ground at C, at 30° with the ground.
Given

Rope AC=20AC = 20 m, ∠ACB=30°\angle ACB = 30°, pole AB⊥AB \perp ground

To find

Height ABAB of the pole

In right △ABC\triangle ABC: sin⁡30°=ABAC\sin 30° = \dfrac{AB}{AC}ABAB opposite, ACAC hypotenuse
12=AB20\dfrac12 = \dfrac{AB}{20}sin⁡30°=12\sin 30° = \dfrac12
AB=10AB = 10 m
Answer

The height of the pole is 10 m.

2 Height of the broken tree

Question 2

Q. 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° 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.

30° A (foot) B (break) C original top AC = 8 m broken part BC AB = ?
Figure, Question 2: the tree AB is broken at B; the top touches the ground at C, 8 m from the foot A, at 30°.
Given

AC=8AC = 8 m, ∠ACB=30°\angle ACB = 30°, standing part ABAB, broken part BCBC

To find

Original height AB+BCAB + BC

In right △BAC\triangle BAC: tan⁡30°=ABAC⇒13=AB8\tan 30° = \dfrac{AB}{AC} \Rightarrow \dfrac{1}{\sqrt3} = \dfrac{AB}{8}
AB=83AB = \dfrac{8}{\sqrt3} mstanding part
cos⁡30°=ACBC⇒32=8BC\cos 30° = \dfrac{AC}{BC} \Rightarrow \dfrac{\sqrt3}{2} = \dfrac{8}{BC}
BC=163BC = \dfrac{16}{\sqrt3} mbroken part (was the upper part of the tree)
Height =AB+BC=83+163=243=2433=83= AB + BC = \dfrac{8}{\sqrt3} + \dfrac{16}{\sqrt3} = \dfrac{24}{\sqrt3} = \dfrac{24\sqrt3}{3} = 8\sqrt3 mrationalise
Answer

The height of the tree was 838\sqrt3 m (about 13.86 m).

3 Length of the two slides

Question 3

Q. 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° 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° to the ground. What should be the length of the slide in each case?

30° 60° A B C D E F BC = 1.5 m EF = 3 m AC = ? DF = ? slide for below 5 years slide for elder children
Figure, Question 3: slide AC with top 1.5 m high at 30°, and the steeper slide DF with top 3 m high at 60°.
Given

Slide 1: BC=1.5BC = 1.5 m, ∠A=30°\angle A = 30°. Slide 2: EF=3EF = 3 m, ∠D=60°\angle D = 60°

To find

Lengths ACAC and DFDF

Slide 1, right △ABC\triangle ABC: sin⁡30°=BCAC⇒12=1.5AC\sin 30° = \dfrac{BC}{AC} \Rightarrow \dfrac12 = \dfrac{1.5}{AC}height and slide length: use sin
AC=3AC = 3 m
Slide 2, right △DEF\triangle DEF: sin⁡60°=EFDF⇒32=3DF\sin 60° = \dfrac{EF}{DF} \Rightarrow \dfrac{\sqrt3}{2} = \dfrac{3}{DF}
DF=63=633=23DF = \dfrac{6}{\sqrt3} = \dfrac{6\sqrt3}{3} = 2\sqrt3 mrationalise
Answer

The slide for children below 5 years should be 3 m long and the slide for elder children 232\sqrt3 m (about 3.46 m) long.

4 Height of the tower, 30 m away

Question 4

Q. 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°. Find the height of the tower.

30° A B C CB = 30 m tower AB = ?
Figure, Question 4: tower AB observed from C, 30 m from its foot, at an angle of elevation of 30°.
Given

CB=30CB = 30 m, ∠ACB=30°\angle ACB = 30°

To find

Height ABAB

In right △ABC\triangle ABC: tan⁡30°=ABBC\tan 30° = \dfrac{AB}{BC}opposite and adjacent: tan
13=AB30⇒AB=303=3033=103\dfrac{1}{\sqrt3} = \dfrac{AB}{30} \Rightarrow AB = \dfrac{30}{\sqrt3} = \dfrac{30\sqrt3}{3} = 10\sqrt3 mrationalise
Answer

The height of the tower is 10310\sqrt3 m (about 17.32 m).

5 Length of the kite string

Question 5

Q. 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°. Find the length of the string, assuming that there is no slack in the string.

60° A (kite) B C AB = 60 m string AC = ?
Figure, Question 5: kite A 60 m above the ground point B; string AC makes 60° with the ground.
Given

AB=60AB = 60 m, ∠ACB=60°\angle ACB = 60°, no slack

To find

Length of string ACAC

In right △ABC\triangle ABC: sin⁡60°=ABAC\sin 60° = \dfrac{AB}{AC}height and string (hypotenuse): sin
32=60AC⇒AC=1203\dfrac{\sqrt3}{2} = \dfrac{60}{AC} \Rightarrow AC = \dfrac{120}{\sqrt3}
AC=12033=403AC = \dfrac{120\sqrt3}{3} = 40\sqrt3 mrationalise
Answer

The length of the string is 40340\sqrt3 m (about 69.28 m).

6 Distance the boy walked

Question 6

Q. 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° to 60° as he walks towards the building. Find the distance he walked towards the building.

30° 60° A E D C distance walked DC = ? 30 m building 1.5 m BE = 1.5 m
Figure, Question 6: the boy’s eye moves from D to C (1.5 m above the ground); E is the point on the building at eye level; A is the top.
Given

Building =30= 30 m, boy’s eye height =1.5= 1.5 m, so AE=30−1.5=28.5AE = 30 – 1.5 = 28.5 m; angles 30°30° at DD and 60°60° at CC

To find

Distance walked DCDC

In right △AED\triangle AED: tan⁡30°=AEDE⇒DE=28.53\tan 30° = \dfrac{AE}{DE} \Rightarrow DE = 28.5\sqrt3 mfirst position
In right △AEC\triangle AEC: tan⁡60°=AECE⇒CE=28.53=28.533=9.53\tan 60° = \dfrac{AE}{CE} \Rightarrow CE = \dfrac{28.5}{\sqrt3} = \dfrac{28.5\sqrt3}{3} = 9.5\sqrt3 msecond position
DC=DE−CE=28.53−9.53=193DC = DE – CE = 28.5\sqrt3 – 9.5\sqrt3 = 19\sqrt3 m
Answer

The boy walked 19319\sqrt3 m (about 32.91 m) towards the building.

7 Height of the transmission tower

Question 7

Q. 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° and 60° respectively. Find the height of the tower.

45° 60° P B C D building BC = 20 m tower CD = ?
Figure, Question 7: building BC = 20 m with tower CD on top; from P the angles of elevation of C and D are 45° and 60°.
Given

BC=20BC = 20 m, ∠BPC=45°\angle BPC = 45°, ∠BPD=60°\angle BPD = 60°

To find

Height of tower CDCD

In right △PBC\triangle PBC: tan⁡45°=BCPB⇒1=20PB\tan 45° = \dfrac{BC}{PB} \Rightarrow 1 = \dfrac{20}{PB}
PB=20PB = 20 m
In right △PBD\triangle PBD: tan⁡60°=BDPB⇒BD=203\tan 60° = \dfrac{BD}{PB} \Rightarrow BD = 20\sqrt3 m
CD=BD−BC=203−20=20(3−1)CD = BD – BC = 20\sqrt3 – 20 = 20(\sqrt3 – 1) m
Answer

The height of the transmission tower is 20(3−1)20(\sqrt3 – 1) m (about 14.64 m).

8 Height of the pedestal

Question 8

Q. 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° and from the same point the angle of elevation of the top of the pedestal is 45°. Find the height of the pedestal.

45° 60° P B C D statue CD = 1.6 m pedestal BC = ?
Figure, Question 8: pedestal BC with a 1.6 m statue CD on top; from P the angles of elevation of D and C are 60° and 45°.
Given

CD=1.6CD = 1.6 m, ∠BPD=60°\angle BPD = 60°, ∠BPC=45°\angle BPC = 45°; let BC=hBC = h m

To find

Height of pedestal hh

In right △PBC\triangle PBC: tan⁡45°=hPB⇒PB=h\tan 45° = \dfrac{h}{PB} \Rightarrow PB = h
In right △PBD\triangle PBD: tan⁡60°=h+1.6PB⇒3=h+1.6h\tan 60° = \dfrac{h + 1.6}{PB} \Rightarrow \sqrt3 = \dfrac{h + 1.6}{h}
h3=h+1.6⇒h(3−1)=1.6h\sqrt3 = h + 1.6 \Rightarrow h(\sqrt3 – 1) = 1.6
h=1.63−1=1.6(3+1)(3−1)(3+1)=1.6(3+1)2=0.8(3+1)h = \dfrac{1.6}{\sqrt3 – 1} = \dfrac{1.6(\sqrt3 + 1)}{(\sqrt3 – 1)(\sqrt3 + 1)} = \dfrac{1.6(\sqrt3 + 1)}{2} = 0.8(\sqrt3 + 1) mrationalise
Answer

The height of the pedestal is 0.8(3+1)0.8(\sqrt3 + 1) m (about 2.19 m).

9 Height of the building from the tower

Question 9

Q. The angle of elevation of the top of a building from the foot of the 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.

30° 60° A B C D tower AB = 50 m building CD = ?
Figure, Question 9: tower AB = 50 m and building CD; from B the elevation of C is 30°, from D the elevation of A is 60°.
Given

AB=50AB = 50 m, ∠ADB=60°\angle ADB = 60°, ∠CBD=30°\angle CBD = 30°

To find

Height of building CDCD

In right △ABD\triangle ABD: tan⁡60°=ABBD⇒3=50BD\tan 60° = \dfrac{AB}{BD} \Rightarrow \sqrt3 = \dfrac{50}{BD}the tower’s height is known, start here
BD=503BD = \dfrac{50}{\sqrt3} m
In right △CDB\triangle CDB: tan⁡30°=CDBD⇒CD=13×503\tan 30° = \dfrac{CD}{BD} \Rightarrow CD = \dfrac{1}{\sqrt3} \times \dfrac{50}{\sqrt3}
CD=503=1623CD = \dfrac{50}{3} = 16\tfrac{2}{3} m
Answer

The height of the building is 162316\tfrac{2}{3} m (about 16.67 m).

10 Two equal poles across an 80 m road

Question 10

Q. 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° and 30°, respectively. Find the height of the poles and the distances of the point from the poles.

60° 30° A B C D P x 80 − x h h
Figure, Question 10: equal poles AB and CD on either side of an 80 m road; from P between them the angles of elevation of A and C are 60° and 30°.
Given

AB=CD=hAB = CD = h, BD=80BD = 80 m, BP=xBP = x, PD=80−xPD = 80 – x; ∠APB=60°\angle APB = 60°, ∠CPD=30°\angle CPD = 30°

To find

hh, BPBP and PDPD

In right △ABP\triangle ABP: tan⁡60°=hx⇒h=x3\tan 60° = \dfrac{h}{x} \Rightarrow h = x\sqrt3(1)
In right △CDP\triangle CDP: tan⁡30°=h80−x⇒h=80−x3\tan 30° = \dfrac{h}{80 – x} \Rightarrow h = \dfrac{80 – x}{\sqrt3}(2)
From (1) and (2): x3=80−x3⇒3x=80−xx\sqrt3 = \dfrac{80 – x}{\sqrt3} \Rightarrow 3x = 80 – xcross-multiply
4x=80⇒x=204x = 80 \Rightarrow x = 20 m, so PD=80−20=60PD = 80 – 20 = 60 m
h=203h = 20\sqrt3 mfrom (1)
Answer

Each pole is 20320\sqrt3 m (about 34.64 m) high. The point is 20 m from the pole with the 60°60° angle and 60 m from the other pole.

11 TV tower and width of the canal

Question 11

Q. 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°. 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° (see Fig. 9.12). Find the height of the tower and the width of the canal.

60° 30° A B C D canal BC = ? CD = 20 m TV tower AB = ?
Figure, Question 11 (NCERT Fig. 9.12): TV tower AB on one bank; C directly opposite on the other bank (60°), D is 20 m farther from C (30°).
Given

CD=20CD = 20 m, ∠ACB=60°\angle ACB = 60°, ∠ADB=30°\angle ADB = 30°; let AB=hAB = h, BC=xBC = x

To find

Height hh and width xx

In right △ABC\triangle ABC: tan⁡60°=hx⇒h=x3\tan 60° = \dfrac{h}{x} \Rightarrow h = x\sqrt3(1)
In right △ABD\triangle ABD: tan⁡30°=hx+20⇒h3=x+20\tan 30° = \dfrac{h}{x + 20} \Rightarrow h\sqrt3 = x + 20(2)
Put (1) in (2): x3⋅3=x+20⇒3x=x+20x\sqrt3 \cdot \sqrt3 = x + 20 \Rightarrow 3x = x + 20
x=10x = 10 m, so h=103h = 10\sqrt3 m
Answer

The height of the TV tower is 10310\sqrt3 m (about 17.32 m) and the width of the canal is 10 m.

12 Height of the cable tower

Question 12

Q. 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°. Determine the height of the tower.

60° 45° A B C D E 7 m CE = AB = 7 m DE = ?
Figure, Question 12: from the top A of the 7 m building AB, the elevation of the top D of the cable tower is 60° and the depression of its foot C is 45°; AE is horizontal.
Given

AB=7AB = 7 m, depression of C=45°C = 45°, elevation of D=60°D = 60°; AE∥BCAE \parallel BC, so CE=AB=7CE = AB = 7 m

To find

Height of tower CDCD

Depression of CC is 45°45°, so ∠ACB=45°\angle ACB = 45° (alternate angles)
In right △ABC\triangle ABC: tan⁡45°=ABBC⇒BC=7\tan 45° = \dfrac{AB}{BC} \Rightarrow BC = 7 m, so AE=7AE = 7 m
In right △AED\triangle AED: tan⁡60°=DEAE⇒DE=73\tan 60° = \dfrac{DE}{AE} \Rightarrow DE = 7\sqrt3 m
CD=CE+DE=7+73=7(3+1)CD = CE + DE = 7 + 7\sqrt3 = 7(\sqrt3 + 1) m
Answer

The height of the cable tower is 7(3+1)7(\sqrt3 + 1) m (about 19.12 m).

13 Distance between two ships

Question 13

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

30° 45° 45° 30° A B C D lighthouse 75 m CD = ?
Figure, Question 13: from the top A of the 75 m lighthouse AB, ships C and D (same side) have angles of depression 45° and 30°.
Given

AB=75AB = 75 m; ∠ACB=45°\angle ACB = 45°, ∠ADB=30°\angle ADB = 30° (alternate angles)

To find

Distance CDCD

In right △ABC\triangle ABC: tan⁡45°=75BC⇒BC=75\tan 45° = \dfrac{75}{BC} \Rightarrow BC = 75 mnearer ship
In right △ABD\triangle ABD: tan⁡30°=75BD⇒BD=753\tan 30° = \dfrac{75}{BD} \Rightarrow BD = 75\sqrt3 mfarther ship
CD=BD−BC=753−75=75(3−1)CD = BD – BC = 75\sqrt3 – 75 = 75(\sqrt3 – 1) m
Answer

The distance between the two ships is 75(3−1)75(\sqrt3 – 1) m (about 54.9 m).

14 Distance travelled by the balloon

Question 14

Q. 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°. After some time, the angle of elevation reduces to 30° (see Fig. 9.13). Find the distance travelled by the balloon during the interval.

30° 60° P Q E 1.2 m R S distance travelled PQ = ? QS = 87 m
Figure, Question 14 (NCERT Fig. 9.13): the girl’s eye E is 1.2 m above the ground; the balloon moves from P to Q at 88.2 m height; R and S are the points at eye level below P and Q.
Given

Height of balloon above the eye PR=QS=88.2−1.2=87PR = QS = 88.2 – 1.2 = 87 m; ∠PER=60°\angle PER = 60°, ∠QES=30°\angle QES = 30°

To find

Distance PQ=RSPQ = RS

In right △PRE\triangle PRE: tan⁡60°=PRER⇒ER=873=8733=293\tan 60° = \dfrac{PR}{ER} \Rightarrow ER = \dfrac{87}{\sqrt3} = \dfrac{87\sqrt3}{3} = 29\sqrt3 m
In right △QSE\triangle QSE: tan⁡30°=QSES⇒ES=873\tan 30° = \dfrac{QS}{ES} \Rightarrow ES = 87\sqrt3 m
PQ=RS=ES−ER=873−293=583PQ = RS = ES – ER = 87\sqrt3 – 29\sqrt3 = 58\sqrt3 mthe balloon moves horizontally, so PQ=RSPQ = RS
Answer

The balloon travelled 58358\sqrt3 m (about 100.46 m) during the interval.

15 Time for the car to reach the tower

Question 15

Q. 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°, 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.

30° 60° A B C D tower h 6 seconds time = ?
Figure, Question 15: from the top A of tower AB the car is seen at D (depression 30°) and six seconds later at C (depression 60°).
Given

Let AB=hAB = h. Depression 30°30° at DD, 60°60° at CC (alternate angles give ∠ADB=30°\angle ADB = 30°, ∠ACB=60°\angle ACB = 60°). DD to CC takes 6 s at uniform speed.

To find

Time from CC to the foot BB

In right △ABD\triangle ABD: tan⁡30°=hBD⇒BD=h3\tan 30° = \dfrac{h}{BD} \Rightarrow BD = h\sqrt3
In right △ABC\triangle ABC: tan⁡60°=hBC⇒BC=h3\tan 60° = \dfrac{h}{BC} \Rightarrow BC = \dfrac{h}{\sqrt3}
DC=BD−BC=h3−h3=3h−h3=2h3DC = BD – BC = h\sqrt3 – \dfrac{h}{\sqrt3} = \dfrac{3h – h}{\sqrt3} = \dfrac{2h}{\sqrt3}covered in 6 s
Speed =2h3÷6=h33= \dfrac{2h}{\sqrt3} \div 6 = \dfrac{h}{3\sqrt3} per second
Time for BC=h3÷h33=h3×33h=3BC = \dfrac{h}{\sqrt3} \div \dfrac{h}{3\sqrt3} = \dfrac{h}{\sqrt3} \times \dfrac{3\sqrt3}{h} = 3 s
Answer

The car will take 3 seconds to reach the foot of the tower. (Shortcut: BCBC is exactly half of DCDC, so it takes half of 6 s.)

16 In-text questions

Section 9.1: “Are they angles of elevation or angles of depression?”

Q. Now, you may identify the lines of sight, and the angles so formed in Fig. 8.3. Are they angles of elevation or angles of depression?

Answer: Fig. 8.3 (from the previous chapter) shows a girl and her mother on the ground looking at a hot air balloon, first at point A and then at point B. Each line of sight runs from an eye on the ground up to the balloon, so the balloon is above the horizontal through the eye and they raise their heads. So both angles are angles of elevation. (In Fig. 9.3, by contrast, the girl on the balcony looks down at the flower pot, so that angle is an angle of depression.)

Example 6: “PD + 8 = BD = PD3\sqrt3 (Why?)”

Q. Since AC = BD and DC = AB = 8 m, we get PD + 8 = BD = PD3\sqrt3. Why?

Answer: AB (the 8 m building) and DC are vertical, and BD and AC are horizontal, so ABDCABDC is a rectangle: AC=BDAC = BD and DC=AB=8DC = AB = 8 m. From tan⁡45°=1\tan 45° = 1, PC=ACPC = AC, and PC=PD+DC=PD+8PC = PD + DC = PD + 8. So PD+8=AC=BDPD + 8 = AC = BD, and from △PBD\triangle PBD, BD=PD3BD = PD\sqrt3.

More for this chapter
NotesImportant QuestionsMCQ / ObjectiveShort Notes