Question

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From a point on the ground, the angle of elevation of the top of a tower is observed to be 60°. From a point 40 m vertically above the first 300 AC point of observation, the angle of elevation of the top of the tower is 30°. Find the height of the tower and its horizontal distance from the point of observation




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Answer 1

${\huge{\sf{\color{red}{★\:An}{\color{orange}{sW}{\color{purple}{er:-}}}}}}$★

• Height of the tower is 60m.

Given:-

• the angle of elevation of the top of a tower is 60°.

• From a point 40 m vertically above the first 300 AC point of observation, the angle of elevation of the top of the tower is 30°.

Solution:-

• Tan 30 = AB/BE

1 /√3 = x / BE

BE = x√3m

• Tan 60 = AC/CD

√3 = x + 40/CD

CD√3 = x + 40

• Since BE is parallel to CD

x√3 x √3 = x + 40

3x = x + 40

3x - x = 40

2x = 40

x = 20

• The height of the tower = x + 40

20 + 40 = 60m

The horizontal distance from the point of observation = x√3 = 20√3m

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Hope it helps buddy : )

Answer 2

{\huge{\sf{\color{red}{★\:An}{\color{orange}{sW}{\color{purple}{er:-}}}}}}★AnsWer:− ★

Height of the tower is 60m.

Given:-

the angle of elevation of the top of a tower is 60°.

From a point 40 m vertically above the first 300 AC point of observation, the angle of elevation of the top of the tower is 30°.

Solution:-

Tan 30 = AB/BE

1 /√3 = x / BE

BE = x√3m

Tan 60 = AC/CD

√3 = x + 40/CD

CD√3 = x + 40

Since BE is parallel to CD

x√3 x √3 = x + 40

3x = x + 40

3x - x = 40

2x = 40

x = 20

The height of the tower = x + 40

20 + 40 = 60m

The horizontal distance from the point of observation = x√3 = 20√3m

-------------------------------------------------------------------------------

Hope it helps buddy : )