Question

6. The angle of elevation of the top of a
building from the foot of the tower is
30° and the angle of elevation of the
top
of the tower from the foot of the
building is 60°. If the tower is 30 m
high, find the height of the building.​

Answer 1

Answer:

Given height of tower CD=30m

0mLet the height of the building, AB=h

0mLet the height of the building, AB=hIn right angled △BDC,

0mLet the height of the building, AB=hIn right angled △BDC,⇒

$\tan(60°) = \frac{BD}{CD}$

⇒

$\sqrt{3 } = \frac{50}{BD}$

$⇒ BD= \frac{50}{ \sqrt{3} } m$

️‍️

In right angled △ABD,

In right angled △ABD,

$⇒ tan30° = \frac{AB}{BD}$

$⇒ \frac{1}{ \sqrt{3} } = \frac{h}{ \frac{50}{ \sqrt{3} } }$

️‍️

$∴ \frac{1}{ \sqrt{3} } = \frac{ \sqrt{3h} }{50}$

$∴ h= \frac{50}{ \sqrt{3} } = 16.66m$

∴ The height of the building is 16.66m.

Answer 2

Answer:

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