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Question
Lakshaman Jhula is located 5 kilometers north-east of the city of Rishikesh in the Indian state of Uttarakhand. The bridge connects the villages of Tapovan to Jonk. Tapovan is in Tehri Garhwal district, on the west bank of the river, while Jonk is in Pauri Garhwal district, on the east bank. Lakshman Jhula is a pedestrian bridge also used by motorbikes. It is a landmark of Rishikesh. A group of Class X students visited Rishikesh in Uttarakhand on a trip. They observed from a point (P) on a river bridge that the angles of depression of opposite banks of the river are 60° and 30° respectively. The height of the bridge is about 18 meters from the river. |
Based on the above information answer the following questions.
- Find the distance PA.
- Find the distance PB
- Find the width AB of the river.
[OR]
Find the height BQ if the angle of the elevation from P to Q be 30°.
Solution
I. sin 60° = `(PC)/(PA)`
⇒ `sqrt(3)/2 = 18/(PA)`
⇒ PA = `12sqrt(3)` m
II. sin 30° = `(PC)/(PB)`
⇒ `1/2 = 18/(PB)`
⇒ PB = 36 m
III. tan 60° = `(PC)/(AC)`
⇒ `sqrt(3) = 18/(AC)`
⇒ AC = `6sqrt(3)` m
tan 30° = `(PC)/(CB)`
⇒ `1/sqrt(3) = 18/(CB)`
⇒ CB = `18sqrt(3)` m
Width AB = AC + CB
= `6sqrt(3) + 18sqrt(3)`
= `24sqrt(3)` m
[OR]
RB = PC = 18 m and PR = CB = `18sqrt(3)` m
tan 30° = `(QR)/(PR)`
⇒ `1/sqrt(3) = (QR)/(18sqrt(3))`
⇒ QR = 18 m
QB = QR + RB
= 18 + 18
= 36 m
Hence height BQ is 36 m.
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