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Chapter 9

Some Applications of Trigonometry

NCERT solutions and explanations for Class 10 Mathematics Chapter 9 Some Applications of Trigonometry — covering angle of elevation, angle of depression, and solving real-world height/distance problems using trigonometric ratios.

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Questions

7
Q1

Define the terms (i) angle of elevation and (ii) angle of depression. What is a line of sight?

(i) Angle of elevation: The angle formed by the line of sight with the horizontal when the object viewed is above the horizontal level (the observer raises their head). (ii) Angle of depression: The angle formed by the line of sight with the horizontal when the object viewed is below the horizontal level (the observer lowers their head). The line of sight is the line drawn from the eye of the observer to the point in the object being viewed.
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Q2

A tower stands vertically on the ground. From a point on the ground, which is 15 m15\text{ m} away from the foot of the tower, the angle of elevation of the top of the tower is found to be 60∘60^\circ. Find the height of the tower.

Let ABAB be the tower and CC be the observation point. In right △ABC\triangle ABC, tan⁡60∘=ABBC\tan 60^\circ = \frac{AB}{BC}. With tan⁡60∘=3\tan 60^\circ = \sqrt{3} and BC=15 mBC = 15\text{ m}, we get 3=AB15\sqrt{3} = \frac{AB}{15}, so AB=153 mAB = 15\sqrt{3}\text{ m}.
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Q3

An electrician has to repair an electric fault on a pole of height 5 m5\text{ m}. She needs to reach a point 1.3 m1.3\text{ m} below the top of the pole. What should be the length of the ladder that she should use which, when inclined at an angle of 60∘60^\circ to the horizontal, would enable her to reach the required position? Also, how far from the foot of the pole should she place the foot of the ladder? (Take 3=1.73\sqrt{3} = 1.73)

The point on the pole is at height 5−1.3=3.7 m5 - 1.3 = 3.7\text{ m}. Using sin⁡60∘=3.7BC\sin 60^\circ = \frac{3.7}{BC} (ladder length BCBC): BC=3.7×23=7.41.73≈4.28 mBC = \frac{3.7 \times 2}{\sqrt{3}} = \frac{7.4}{1.73} \approx 4.28\text{ m}. Using cot⁡60∘=DC3.7\cot 60^\circ = \frac{DC}{3.7} (distance DCDC): DC=3.73=3.71.73≈2.14 mDC = \frac{3.7}{\sqrt{3}} = \frac{3.7}{1.73} \approx 2.14\text{ m}.
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Q4

An observer 1.5 m1.5\text{ m} tall is 28.5 m28.5\text{ m} away from a chimney. The angle of elevation of the top of the chimney from her eyes is 45∘45^\circ. What is the height of the chimney?

Let the chimney be ABAB and observer CDCD of height 1.5 m1.5\text{ m}. In right △ADE\triangle ADE, tan⁡45∘=AEDE=AE28.5\tan 45^\circ = \frac{AE}{DE} = \frac{AE}{28.5}. Since tan⁡45∘=1\tan 45^\circ = 1, AE=28.5 mAE = 28.5\text{ m}. Height of chimney AB=AE+BE=28.5+1.5=30 mAB = AE + BE = 28.5 + 1.5 = 30\text{ m}.
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Q5

From a point PP on the ground, the angle of elevation of the top of a 10 m10\text{ m} tall building is 30∘30^\circ. A flag is hoisted at the top of the building and the angle of elevation of the top of the flagstaff from PP is 45∘45^\circ. Find the length of the flagstaff and the distance of the building from point PP. (Take 3=1.732\sqrt{3} = 1.732)

Let building height AB=10 mAB = 10\text{ m}, flagstaff BD=x mBD = x\text{ m}. In △PAB\triangle PAB, tan⁡30∘=10AP\tan 30^\circ = \frac{10}{AP}, so AP=103≈17.32 mAP = 10\sqrt{3} \approx 17.32\text{ m}. In △PAD\triangle PAD, tan⁡45∘=10+xAP=1\tan 45^\circ = \frac{10 + x}{AP} = 1, so 10+x=10310 + x = 10\sqrt{3}, giving x=10(3−1)≈7.32 mx = 10(\sqrt{3} - 1) \approx 7.32\text{ m}. Distance AP=103≈17.32 mAP = 10\sqrt{3} \approx 17.32\text{ m}.
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Q6

The shadow of a tower standing on a level ground is found to be 40 m40\text{ m} longer when the Sun's altitude is 30∘30^\circ than when it is 60∘60^\circ. Find the height of the tower.

Let tower height AB=h mAB = h\text{ m}, shadow at 60∘60^\circ altitude be BC=x mBC = x\text{ m}. Shadow at 30∘30^\circ is BD=(x+40) mBD = (x + 40)\text{ m}. From △ABC\triangle ABC, tan⁡60∘=hx\tan 60^\circ = \frac{h}{x} gives h=x3h = x\sqrt{3}. From △ABD\triangle ABD, tan⁡30∘=hx+40=13\tan 30^\circ = \frac{h}{x + 40} = \frac{1}{\sqrt{3}}. Substituting h=x3h = x\sqrt{3}: x3x+40=13\frac{x\sqrt{3}}{x + 40} = \frac{1}{\sqrt{3}}, so 3x=x+403x = x + 40, giving x=20x = 20 and h=203 mh = 20\sqrt{3}\text{ m}.
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Q7

From a point on a bridge across a river, the angles of depression of the banks on opposite sides of the river are 30∘30^\circ and 45∘45^\circ, respectively. If the bridge is at a height of 3 m3\text{ m} from the banks, find the width of the river.

Let bridge point PP at height DP=3 mDP = 3\text{ m}. In right △APD\triangle APD, tan⁡30∘=3AD\tan 30^\circ = \frac{3}{AD}, so AD=33 mAD = 3\sqrt{3}\text{ m}. In right △BPD\triangle BPD, tan⁡45∘=3BD\tan 45^\circ = \frac{3}{BD}, so BD=3 mBD = 3\text{ m}. Width AB=AD+DB=33+3=3(3+1) mAB = AD + DB = 3\sqrt{3} + 3 = 3(\sqrt{3} + 1)\text{ m}.
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