Watch from the top of a $80 m$ high hill, the angle of elevation of the top of a tower is $30$ and the angle of depression of the base of the tower is $45$ . Find the height of the tower.

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Solution

Let height of tower $= h$
$A E = 80 = C D =$ Height of hill$
Let $A D = x$
Now, $tan 30 ° = \frac{1}{\sqrt{3}} = \frac{h - 80}{x}$
So, $x = \sqrt{3} (h - 80)$
and $tan 45 ° = 1 = \frac{80}{x} ⟶ x = 80$
So, $80 = \sqrt{3} (h - 80) h = \frac{80 (1 + \sqrt{3})}{\sqrt{3}} \approx 126.18 m$

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The angle of elevation of the top of a hill at the foot of a tower is $60^{\circ}$ and the angle of depression from the top of the tower to the foot of the hill is $30^{\circ}$ . If the tower is 50 m high, find the height of the hill.

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Watch from the top of a 80 m high hill, the angle of elevation of the top of a tower is 30 and the angle of depression of the base of the tower is 45. Find the height of the tower.

Let height of tower =hAE=80=CD= Height of hill$Let AD=xNow, tan30°=1√3=h−80xSo, x=√3(h−80)and tan45°=1=80x⟶x=80So, 80=√3(h−80)h=80(1+√3)√3≈126.18m

Watch from the top of a $80 m$ high hill, the angle of elevation of the top of a tower is $30$ and the angle of depression of the base of the tower is $45$ . Find the height of the tower.

Let height of tower $= h$
$A E = 80 = C D =$ Height of hill$
Let $A D = x$
Now, $tan 30 ° = \frac{1}{\sqrt{3}} = \frac{h - 80}{x}$
So, $x = \sqrt{3} (h - 80)$
and $tan 45 ° = 1 = \frac{80}{x} ⟶ x = 80$
So, $80 = \sqrt{3} (h - 80) h = \frac{80 (1 + \sqrt{3})}{\sqrt{3}} \approx 126.18 m$