Question

ғʀᴏᴍ ᴛʜᴇ ᴛᴏᴘ ᴏғ ᴀ ᴛᴏᴡᴇʀ, ᴀ ᴍᴀɴ ғɪɴᴅs ᴛʜᴀᴛ ᴛʜᴇ ᴀɴɢʟᴇ ᴏғ ᴅᴇᴘʀᴇssɪᴏɴ ᴏғ ᴀ ᴄᴀʀ ᴏɴ ᴛʜᴇ ɢʀᴏᴜɴᴅ ɪs 30°.if ᴛʜᴇ ᴄᴀʀ ɪs ᴀ ᴅɪsᴛᴀɴᴄᴇ ᴏғ 40 m ғʀᴏᴍ ᴛʜᴇ ᴛᴏᴡᴇʀ ғɪɴᴅ ᴛʜᴇ ʜᴇɪɢʜᴛ ᴏғ ᴀʟᴀ ᴛᴏᴡᴇʀ​

Answer 1

$\sf{\huge{\underline{\red{Given :-}}}}$

• A man is on the top of tower .

• Angle of depression of car on the ground is 30°.

• The distance from tower is 30° from car.

$\sf{\huge{\underline{\pink{To \:Find :-}}}}$

• The height of the tower .

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

We have ,

• The distance between car & tower is 40m.

• The Angle of depression is 30°.

Take a right triangle ABC,

We , know tan θ = $\dfrac{perpendicular}{base}$

➝ tan 30 = $\dfrac{perpendicular}{40}$

We know, $\dfrac{1}{ \sqrt{3} }$

➝ $\dfrac{1}{ \sqrt{3} }$ = $\dfrac{perpendicular}{40}$

➝ $\dfrac{1}{ \sqrt{3} }$ × 40 = perpendicular

➝ Perpendicuar = $\frac{40}{ \sqrt{3} }$

➝ Perpendicuar = 40× 1.733.

➝ Perpendicular = 69.32m

Hence,The height of the tower is 69.32m.

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Answer 2

$\huge{ { \boxed{ \mathtt { \red{✯AnSwEr✯\ }}}}}$

We have ,

The distance between car & tower is 40m.

The Angle of depression is 30°.

Take a right triangle ABC,

We , know tan θ = \dfrac{perpendicular}{base}

base

perpendicular

➝ tan 30 = \dfrac{perpendicular}{40}

40

perpendicular

We know, \dfrac{1}{ \sqrt{3} }

3

1

➝ \dfrac{1}{ \sqrt{3} }

3

1

= \dfrac{perpendicular}{40}

40

perpendicular

➝ \dfrac{1}{ \sqrt{3} }

3

1

× 40 = perpendicular

➝ Perpendicuar = \frac{40}{ \sqrt{3} }

3

40

➝ Perpendicuar = 40× 1.733.

➝ Perpendicular = 69.32m

Hence,The height of the tower is 69.32m.