Question

A tower subtends an angle of 60° at a point on the same level as the foot of the
At a second point 20 m above the first, the angle of depression
is 45°. Find the height of the tower.​

Answer 1

now ...

tan45°=(h-20)/X

h-20=X

and ...

tan60°=h/X

=>(√3)x=h

=>√3(h-20)=h

=>h√3-h=20√3

=>h(√3-1)=20√3

$h = \frac{20 \sqrt{3} }{ \sqrt{3} - 1} \\ = > h = \frac{20 \sqrt{3} ( \sqrt{3} + 1) }{3 - 1} \\ = > h = 10 \sqrt{3} ( \sqrt{3} + 1) \\ = > h = 30 + 10 \sqrt{3} \\ = > h = 30 + 17.32 \\ = > h = 47.32 \: \: metre$

Answer 2

SOLUTION:-

Let the AB be the height be the tower.

Let D be the point where the tower subtends angle of 60°

Let C be the point where such that CD= 20m.

From the angle of depression subtended at the foot of the tower is 45°

In ∆CDB,

$tan45 \degree = \frac{H}{BD} \\ \\ = >BD = h \: cot45 \degree \\ \\ = > BD = 20 \times 1 \\ = > BD = 20m...........(1)$

In ∆ADB,

$tan60 \degree = \frac{AB}{BD} \\ \\ = > \sqrt{ 3} = \frac{AB}{20} \\ \\ = > AB = 20 \sqrt{ 3} m$

=) 20× 1.732

=) 34.64m

Thus, 34.64m required Height of the tower.

Hope it helps ☺️