Question

plz answer it....I ll be thankful​

Answer 1

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see in the attachment....and please Bhai mark me best...and give thanks ....yo

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Answer 2

Answer to Qn 36:

(i)  sin B  =  (√ 4b² - a² ) / 2b

(ii) sin ∠BAD  = a / 2b

Step-by-step explanation:

ΔABC is isosceles

so, the perpendicular from A divides BC into two equal parts.

⇒BD = BC / 2 = $\frac{a}{2}$

Δle ABD is right angled,

by pythagoras theorem,

AB² = AD² + BD²

AD² = AB² - BD²

       = b² - $(\frac{a}{2} )^{2}$          

       =  (4b² - a² ) / 4

∴ AD = (√ 4b² - a² ) / 2  

i) sin B =  AD / AB

∴ sin B  =  (√ 4b² - a² ) / 2b

ii) sin ∠BAD = BD / AB

∴ sin ∠BAD  = a / 2b