Question

$\huge\bigstar\huge\tt\underline\red{QUESTION}\bigstar$

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 above attachment)Find the height of the tower and the width of the canal.





SOLVE PLEASE
$\huge\bigstar\huge\tt\underline\red{COPIED ANSWERS NOT ALLOWED}\bigstar$
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Answer 1

Answer:

$\huge\mathfrak{Bonjour!!}</p><p>$

$\huge\fcolorbox{black}{aqua}{Solution:-} </p><p>$

✒ 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.

__________________________

í հօթҽ Եհís հҽlթs!

Answer 2

$let \: power \: be \: = ab \\ so \: height \: of \: tower \: = ab$

$given \: that \: from \: a \: point \: on \: the \: other \: bank \: directly \: opposite \: the \: tower \: the \: angle \: of \: elevation \: of \: the \: tower \: is \: 60$

hence /_ ACB=30°.

$also \: angle \: of \: elevation \: from \: point \: d \: to \: top \: of \: the \: tower \: = 30$

so,/_ ADB =30°

Also DC is 20 m.

$we \: need \: to \: find \: height \: of \: tower \: ab \: and \: width \: of \: canal \: bc \\ since \: tower \: is \: vertical \: to \: ground \:$

so,/_ ABD =90°

in right angle triangle ACB,

tan C= side opposite to angle C

side adjacent to angle C

tan 60°=AB

BC

$\sqrt{3} \: = \: ab \: with \: bc \:$

$\sqrt{3} \: bc \: = ab \\ ab = \: \sqrt{3 \: bc}$

In a right angle triangle ADB,

tan D = side opposite to angle D

side and jacent to angle D

tan 30° = AB

BD

1

√3 = AB

BD

BD

√3 =AB

AB = BD

√3

From 1 and 2

√3BC=BD

√3

√3 × √3 BC = BD

3 BC=BD

3BC =BC + CD

3 BC = BC +20

3BC -BC =20

2BC =20

BC = 20

2

BC =10m

Hence the width of the canal is 10 m

From =1

AB = √3 BC

AB = √3 × 10

AB = 10 √3m

hence height of tower =

$10 \: \sqrt{3} \: m$