Question

$<h2 > question$



$<h5 > circle$


⊙ Please solve the two questions of the attachment.


⊙ Especially @ Sidhartaoo77





$<marquee > prabhudutt$

Answer 1

Answer:

41°

Step-by-step explanation:

(11)

∴ Let ∠ABD = ∠ACD = y.

In ΔACB,

⇒ x + (y+ 30) + 90° = 180°

⇒ x + y + 30 + 90 = 180

⇒ x + y = 60°.

In ΔECB,

∠E + ∠C + ∠B = 180°

⇒ 22 + (90 + y) + (30 + y) = 180°

⇒ 22 + 90 + y + 30 + y = 180

⇒ 142 + 2y = 180

⇒ 2y = 38

⇒ y = 19

Now,

⇒ x + y = 60°

⇒ x + 19 = 60

⇒ x = 60 - 19

⇒ x = 41°.

Hope it helps!

Answer 2

Hi there !
Here's the answer:

•°•°•°•°•°<><><<><>><><>°•°•°•°•°•°

11.

Construction: JOIN A TO D

Let <ABD = $\alpha$

Since, <ACD also lies in the Sam segment ,
=> <ACD = $\alpha$

In ∆ACB,
x + ($\alpha$+30°) + 90° = 180°

=> x + $\alpha$ = 180° - 120°
=> x + $\alpha$ = 60° --------[1]

In ∆ECB,
<E + <C + <B = 180°

=> 22° + (90°+ $\alpha$)+(30° + $\alpha$) = 180°
=> 2$\alpha$ = 180°-142°
=> 2$\alpha$ = 38°
=> $\alpha$ = 19°

Substitute in [1]
x + 19° = 60°
=> x = 60° - 19°
=> x = 41°

This answer x= 41° is in option - (B)
Option (B) is the correct answer

•°•°•°•°•°<><><<><>><><>°•°•°•°•°•°

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