Question

Two men standing on opposite sides of a tower measure the angles of elevation of the top of the tower is 30 degree and 60 degree respectively. If the height of the tower is 100 m, then find the distance between the two men.​

Answer 1

Answer:

hope it helps

Step-by-step explanation:

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Answer 2

Answer:

Answer:

$80\sqrt{3}$

Step-by-step explanation:

Given :Two men standing on opposite sides of a tower measure the angles of elevation of the top of the tower is 30 degree and 60 degree respectively.

To Find: If the height of the tower is 20 m, then find the distance between the two men.

Solution :

Refer the attached figure:

AC (Height of tower )= 20 m

In ΔABC

∠ABC = 30°

perpendicular = AC= 20 m

Base = BC

To find BC we will use trigonometric ratios

$tan\theta=\frac{perpendicular}{base}tanθ= </p><p>base \: </p><p>perpendicular$

tan 30 ^{\circ}=\frac{AC}{BC}tan30

∘

=

BC

AC

$</p><p>\frac{1}{\sqrt{3}} =\frac{20}{BC} </p><p>3</p><p> </p><p> </p><p>1</p><p> </p><p> = </p><p>BC</p><p>20$

$BC =20*\sqrt{3}BC=20∗ </p><p>3</p><p>$

InΔADC

∠ADC = 60°

perpendicular = AC= 20 m

Base = DC

To find DC we will use trigonometric ratios

$tan\theta=\frac{perpendicular}{base}tanθ=$

base

perpendicular

$</p><p>tan 60 ^{\circ}=\frac{AC}{DC}tan60 </p><p>∘</p><p> = </p><p>DC</p><p>AC$

$sqrt{3}=\frac{20}{DC} </p><p>3</p><p> </p><p> = </p><p>DC</p><p>20</p><p> </p><p> </p><p></p><p>DC =\frac{20}{\sqrt{3}}DC= </p><p>3</p><p> </p><p> </p><p>20$

$an 60 ^{\circ}=\frac{AC}{DC}tan60 </p><p>∘</p><p> = </p><p>DC</p><p>AC</p><p> </p><p> </p><p></p><p>\sqrt{3}=\frac{20}{DC} </p><p>3</p><p> </p><p> = </p><p>DC</p><p>20</p><p> </p><p> </p><p></p><p>DC =\frac{20}{\sqrt{3}}DC= </p><p>3</p><p> </p><p> </p><p>20$

The distance between two men = BC+CD

$=20\sqrt{3}+\frac{20}{\sqrt{3}}20 </p><p>3</p><p> </p><p> + </p><p>3</p><p> </p><p> </p><p>20</p><p> </p><p> </p><p></p><p>=80\sqrt{3}$

Hence the distance between the two men is

$80\sqrt{3}$