Question

Can anyone solve qno.35 .plss it a urgent.i will mark him or her as brainlist

Answer 1

Answer:

20√3(√3 + 1) m

Step-by-step explanation:

35.

Let BD be the building of height 20 m and AE be the momentum.

From figure, BD = CE = 20 m.

(i) In ΔBED:

tan 15° = BD/DE

⇒ (√3 - 1)/(1 + √3) = 20/DE

⇒ DE(√3 - 1) = 20(√3 + 1)

⇒ DE = 20(√3 + 1)/(√3 - 1)

⇒ DE = 20(√3 + 1)/(√3 - 1) * [(√3 + 1)/(√3 + 1)]

⇒ DE = 20(√3 + 1)²/2

⇒ DE = 20(3 + 1 + 2√3)/2

⇒ DE = 20(4 + 2√3)/2

⇒ DE = 10(4 + 2√3)

⇒ DE = 40 + 20√3

⇒ DE = 20(2 + √3)

Now,

⇒ DE = BC = 20(2 + √3)

(ii) In ΔABC:

tan 45° = AC/BC

⇒ 1 = AC/BC

⇒ 1 = AC/20(2 + √3)

⇒ AC = 20(2 + √3)

∴ Height of the monument(AE) = AC + CE

= 20(2 + √3) + 20

= 20(3 + √3)

= 20√3(√3 + 1) m.

∴ Height of the monument is: 20√3(√3 + 1) m.

Hope it helps!

Answer 2

HELLO DEAR FRIEND,

YOUR ANSWER

⏬⏬⏬⏬⏬

$\large \bf \boxed{(60 + 20 \sqrt{3})meter }$

⏬⏬⏬⏬⏬

20√3(√3 + 1 ) meter

EXPLANATION IN THE ATTACHMENT.

HOPE IT HELPS