Question

From the top of a 10m high building the angle of alevation of the top of a tower is 60 and the angle of depression of its foot is 45 determine the height of tower

Answer 1

HEY BUDDY
HERE'S YOUR ANSWER
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Answer 2

Height of tower is 10(3√+1)m

Step-by-step explanation:

Given:

Let AB be the building of height 10 m and EC be the height of tower.

A is the point from where elevation of tower is 60° and the angle of depression of its foot is 45°

EC = DE + CD

also, CD = AB = 10 m. and BC = AD

In right angled ΔABC,

tan 45° =$\frac{opposite}{adjacent}$

tan 45° =$\frac{AB}{BC}$

1 =$\frac{10}{BC}$                             [∵ tan 45° = 1]

BC = 10 = AD

Also in  right angled ΔADE

tan 60° = $\frac{DE}{AD}$

$\sqrt{3} = \frac{DE}{10}$                      [$tan60 = \sqrt{3}$]

$DE=10\sqrt{3} m$

Height of the tower = EC = DE +CD

                                           =$(10\sqrt{3}+10 )m$

                                           =$10(\sqrt{3}+1 )m$