Question

a man 1.75m tall looks at a dove on the top of a tree with an angle of elevation of 60o. if the distance between the man's eyes and the top of the tree is 30m, what is the height of the tree from the ground?.

Answer 1

Let AB be the man of height 1.75 m and CD be the total height of the tree on whose top the dove is sitting.

Now,

This forms a Right angled triangle AEC and an imaginary rectangle ABDE.

(As shown in Figure)

Since,

ABDE is rectangle.

Thus,

AB = ED = 1. 75 m

Also,

In Triangle AEC :-

$\sin \theta = \frac{ce}{ac} \\ \\ \\ \sin(60) = \frac{ce}{30} \\ \\ \\ \frac{ce}{30} = \frac{ \sqrt{3} }{2} \\ \\ \\ce = 15 \sqrt{3} \\ \\ \\ce = 15 \times 1.732 \\ \\ \\ce = 25.98$

Hence,

Height of Tree = CE+DE

Height will be

$= 25.98 + 1.75 \\ \\ \\ = 27.73m$

Answer 2

Solution :-

Let AB be the man of height 1.75 m.

Let AB be the man of height 1.75 m.Let CD be the total height of the tree on whose top the dove is sitting.

Now,

This forms a Right angled triangle AEC and an imaginary rectangle ABDE.

Since,

ABDE is rectangle.

Thus,

AB = ED = 1.75 m

Also,

In Triangle AEC,

sin θ = ce / ac

→ sin (60) = ce / 30

→ ce / 30 = √3 / 2

→ ce = 15 √3

→ ce = 15 × 1.732

→ ce = 28.98

Hence,

Height of Tree = CE + DE

→ 28.98 + 1.75

→ 30.73 m