Question

11. A TV tower stands vertically on a bank
of a canal. From a point on the other
bank directly opposite the tower, the
angle of elevation of the top of the
tower is 60°. From another point 20 m
away from this point on the line joing
this point to the foot of the tower, the
angle of elevation of the top of the
tower is 30° (see Fig. 9.12). Find the
height of the tower and the width of
the canal.
​

Answer 1

Answer:

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Answer 2

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$\huge\fcolorbox{black}{aqua}{Solution:-}$

✒ ᴀɴsᴡᴇʀ:-

⛄ ★ ʜᴇɪɢʜᴛ ᴏғ ᴛʜᴇ ᴛᴏᴡᴇʀ:- 10√3 ᴍ.

★ ᴡɪᴅᴛʜ ᴏғ ᴛʜᴇ ᴄᴀɴᴀʟ:- 10 ᴍ.

✒ sᴛᴇᴘ-ʙʏ-sᴛᴇᴘ ᴇxᴘʟᴀɴᴀᴛɪᴏɴ:-

⛄ ʟᴇᴛ ʜᴇɪɢʜᴛ ᴏғ ᴛʜᴇ ᴛᴏᴡᴇʀ ʙᴇ ʜ ᴍᴇᴛʀᴇs ᴀɴᴅ ᴡɪᴅᴛʜ ᴏғ ᴛʜᴇ ᴄᴀɴᴀʟ ʙᴇ x ᴍᴇᴛʀᴇs, sᴏ ᴀʙ = ʜ ᴍ ᴀɴᴅ ʙᴄ = x ᴍ.

ɴᴏᴡ, ɪɴ ∆ᴀʙᴄ, ᴡᴇ ʜᴀᴠᴇ

ᴛᴀɴ 60° = ʜ/x

=> √3 = ʜ/x

=> ʜ = √3x .....→ (ɪ)

ɴᴏᴡ, ɪɴ ∆ᴀʙᴅ, ᴡᴇ ʜᴀᴠᴇ

ᴛᴀɴ 30° = ᴀʙ/ᴅʙ

=> 1/√3 = ʜ/20 + x

=> 20 + x = √3ʜ ......→ (ɪɪ)

ғʀᴏᴍ (ɪ) ᴀɴᴅ (ɪɪ), ᴡᴇ ʜᴀᴠᴇ

20 + x = √3 × √3x

=> 20 + x = 3x

=> 20 = 3x - x = 2x

=> x = 20/2

=> x = 10 ᴍ.

ɴᴏᴡ, ᴘᴜᴛᴛɪɴɢ ᴛʜᴇ ᴠᴀʟᴜᴇ ᴏғ x ɪɴ ᴇϙᴜᴀᴛɪᴏɴ (ɪ), ᴡᴇ ʜᴀᴠᴇ

ʜ= √3 × 10 = 10√3

=> ʜ = 10√3 ᴍ.

↪ ʜᴇɴᴄᴇ, ʜᴇɪɢʜᴛ ᴏғ ᴛʜᴇ ᴛᴏᴡᴇʀ ɪs 10√3 ᴍ ᴀɴᴅ ᴡɪᴅᴛʜ ᴏғ ᴛʜᴇ ᴄᴀɴᴀʟ ɪs 10 ᴍ.

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