Question

a woman 1.65. tall stood 50m away from the foot of a tower,and observe that the angle of elevation of the top of the tower to be 50⁰ what is the height of the tower​

Answer 1

$\sf\huge\underline{\star Solution:-}$

$\rightarrow$Let the women's height be AE and distance between the women and tower be AC.

Also let the height of tower be BC.

$\rightarrow$Now, clearly it is forming a triangle.

So, in triangle ABC,

$\rightarrow$$\sf{Tan50° \:= \: \frac{BC}{AC}}$

$\rightarrow$$\sf{1.19 \:= \: \frac{BC}{50}}$

$\rightarrow$$\sf{1.19 ×50\:= \: BC}$

$\rightarrow$$\sf{59.5m\:= \: BC}$

$\rightarrow$Hence, BC = 59.5m

So, BD(total height of the tower)$\sf{ = \:BC+CD}$

$\sf{= \:59.5+1.65}$

$\sf{=\: 61.15m}$

Therefore, total height of the tower =$\sf\purple{ 61.15m.}$

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Hope it helps you:)

Answer 2

Answer:

61.24 m to the nearest hundredth.

Step-by-step explanation:

tan 50 = H / 50  

H =  50 tan 50

Height of the tower = H + height of the woman

= 50 tan 50 + 1.65

= 61.2377 m