Question

A vertical tower stands on the ground and is surmounted by a flag staff of height 5m from a point on the ground,the angle of elevation of the bottom of the flag is 45 and the top of the flag staff is 60.Find the height of the tower.answer fast I will mark brainliest
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Answer 1

Given:

A vertical tower (AB)

AB is surmounted by a flagstaff of height 5m. (AD)

Angle of elevation of the bottom of the flagstaff is 45° (∠DCB)

Angle of elevation of the top of the flagstaff is 60° (∠ACB)

In ΔABC

∠B = 90°

Therefore:

$\rm \Longrightarrow tan \theta = \dfrac{Opposite \ Side}{Adjacent \ Side}$

$\rm \Longrightarrow tan \theta = \dfrac{AB}{BC}$

$\rm \Longrightarrow tan 60^\circ = \dfrac{AD + DB}{BC}$

$\rm \Longrightarrow \sqrt{3} = \dfrac{5 + DB}{BC}$

$\rm \Longrightarrow BC = \dfrac{5 + x}{\sqrt{3}}$  

Let this be Equation(1).

In ΔDBC

∠B = 90°

Therefore:

$\rm \Longrightarrow tan \theta = \dfrac{Opposite \ Side}{Adjacent \ Side}$

$\rm \Longrightarrow tan \theta = \dfrac{DB}{BC}$

$\rm \Longrightarrow tan45^\circ = \dfrac{x}{BC}$

$\rm \Longrightarrow 1 = \dfrac{x}{BC}$

$\rm \Longrightarrow BC= x$

Let this be Equation(2).

From Eq(1) and Eq(2) we get:

$\rm \Longrightarrow x = \dfrac{5 + x}{\sqrt{3}}$

$\rm \Longrightarrow \sqrt{3}x = 5 + x$

$\rm \Longrightarrow \sqrt{3}x - x = 5$

$\rm \Longrightarrow x(\sqrt{3} - 1) = 5$

$\rm \Longrightarrow x = \dfrac{5}{\sqrt{3} - 1}$

$\rm \Longrightarrow x = \dfrac{5}{\sqrt{3} - 1} \times \dfrac{\sqrt{3} + 1}{\sqrt{3} + 1}$

$\rm \Longrightarrow x = \dfrac{5\sqrt{3} + 5}{\big(\sqrt{3}\big)^2 - (1)^2}$

$\rm \Longrightarrow x = \dfrac{5\sqrt{3} + 5}{3 - 1}$

$\rm \Longrightarrow x = \dfrac{5(\sqrt{3} + 1)}{2}$

$\rm \Longrightarrow x = \dfrac{5(1.732 + 1)}{2}$

$\rm \Longrightarrow x = \dfrac{5(2.732)}{2}$

$\rm \Longrightarrow x = \dfrac{13.66}{2}$

$\rm \Longrightarrow x = 6.83m$

Therefore, the height of the tower is 6.83 meters.