ncertmate
Open navigation menu
Class 10
Subjects
Mathematics Science Social Science Civics Social Science Economics Social Science Geography Social Science History
Search
Chapter 9 · Question 7

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.

Back to Chapter

Answer

Direct Answer

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}.

Simple Explanation

The river is 3(3+1)3(\sqrt{3} + 1) metres wide (about 8.2 m8.2\text{ m}). The bridge point divides the river into two parts whose distances are found using the tangent ratios of the given depression angles.

Exam-Ready Structure

Let AA and BB be opposite banks, PP be a point on the bridge, and DD be the foot of the perpendicular from PP to the river surface. Given PD=3 mPD = 3\text{ m}. In right △APD\triangle APD, ∠A=30∘\angle A = 30^\circ (angle of depression from PP to bank AA). Using tan⁡30∘=PDAD\tan 30^\circ = \frac{PD}{AD}: 13=3AD\frac{1}{\sqrt{3}} = \frac{3}{AD}, so AD=33 mAD = 3\sqrt{3}\text{ m}. In right △BPD\triangle BPD, ∠B=45∘\angle B = 45^\circ (angle of depression from PP to bank BB). Using tan⁡45∘=PDBD\tan 45^\circ = \frac{PD}{BD}: 1=3BD1 = \frac{3}{BD}, so BD=3 mBD = 3\text{ m}. Therefore, the width of the river AB=AD+BD=33+3=3(3+1) mAB = AD + BD = 3\sqrt{3} + 3 = 3(\sqrt{3} + 1)\text{ m}. This is approximately 3×2.732=8.196 m3 \times 2.732 = 8.196\text{ m}.

Key Points

  • Angles of depression to two banks are 30∘30^\circ and 45∘45^\circ
  • Bridge height =3 m= 3\text{ m} above banks
  • From tan⁡30∘\tan 30^\circ: AD=33 mAD = 3\sqrt{3}\text{ m}
  • From tan⁡45∘\tan 45^\circ: BD=3 mBD = 3\text{ m}
  • Total width =3(3+1) m= 3(\sqrt{3} + 1)\text{ m}

Related Questions

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.

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?

Report a mistake