Question

To find the width of the river, a man observes the top of a tower on the opposite bank

making an angle of elevation of 61°. When he moves 50m backward from bank and observes

the same top of the tower, his line of vision makes an angle of elevation of 35°. Find the

height of the tower and width of the river. (tan61° = 1.8, tan35° = 0.7)

Answer 1

Answer:

Step-by-step explanation:

Answer 2

Answer:

Width of river is 31.8 m and Height of tower is 57.2 m

Step-by-step explanation:

Given: ∠ACB = 61° and  ∠ADB = 35°  and CD = 50 m

           tan 61° = 1.8   and tan 35° = 0.7

To find: Height of Tower and width of river

Height of Tower = AB and Width of River = CB

In ΔACB

$tan\,61^{\circ}=\frac{AB}{CB}$

$1.8=\frac{AB}{CB}$

$AB=1.8CB$ ..........(1)

In ΔACB

$tan\,35^{\circ}=\frac{AB}{CD}$

$0.7=\frac{AB}{CB+CD}$

$0.7=\frac{1.8CB}{CB+50}$    from (1)

$0.7(CB+50)=1.8CB$

$0.7CB+35=1.8CB$

$1.8CB-0.7CB=35$

$1.1CB=35$

CB = 31.8 m

put this value in (1)

AB=1.8 × 31.8 = 57.24 = 57.2 m

Therefore, Width of river is 31.8 m and Height of tower is 57.2 m