Q) A person standing on the bank of a river observes that the angle of elevation of the top of a tower on the opposite bank is 60°. When he moves 30 m away from the bank, he finds the angle of elevation to be 30°. Find the height of the tower and width of the river. (Take √3 = 1.732)

Ans: Let’s start with the diagram for this question:

A person standing on the bank of a river observes that the angle of elevation of the top of a tower on the opposite bank is 60°. When he moves 30 m away from the bank, he finds the angle of elevation to be 30°. Find the height of the tower and width of the river. (Take √3 = 1.732) 10thBoard CBSE

Here, AB is the tower of height H and AC be the river of width D m.

Step 1: Let’s start from Δ ABC,  tan C = tan 60° = \frac{AB}{AC}

∴  √3 = \frac{H}{D}

∴ H = D√3 ……………. (i)

Step 2: Next, we take Δ ABD, tan 30 = \frac{AB}{AD}

\frac{1}{\sqrt 3} = \frac{H}{30 + D}

∴ 30 + D = H √3 …… (ii)

Step 3: By solving equations (i) and (ii), we get:

∴  30 + D = (D √3) √3

∴ 30 + D = 3 D

∴ 30 = 2 D

D = 15 m

Step 4: 

From equation (i), we have H = D √3 

∴ H = 15 √3 = 15 x 1.732 

H = 25.98 m 

Therefore, height of the tower is 25.98 m and width of the river is 15 m.

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