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Question : 79 of 160
Marks:
+1,
-0
Solution:
‌∠C=90∘,‌ . . . (i0
‌∵‌‌cos‌c=‌=cos‌90∘‌⇒‌‌‌=0‌a2+b2−c2=0‌⇒‌‌c2=a2+b2From Eq. (i), we get
‌=‌ ∵‌=‌=‌=R‌ (constant) ‌ =‌| (Rsin‌A)2−(Rsin‌B)2 |
| (Rsin‌C)2 |
=‌| sin‌2A−sin‌2B |
| sin‌290∘ |
(∵sin‌90∘=1) ‌=(sin‌A+sin‌B)(sin‌A−sin‌B)‌=2⋅sin‌‌⋅cos‌⋅2‌cos‌‌=2sin‌‌⋅cos‌⋅2‌cos‌⋅sin‌‌‌‌‌‌‌∵A+B=π−C‌=2⋅2⋅‌⋅‌⋅sin‌‌⋅‌⋅cos‌‌=2⋅sin‌(‌)⋅cos(‌) ‌‌∵sin‌2A=2sin‌A⋅cos‌A‌=sin‌[2×‌]‌=sin‌(A−B)‌‌ Hence, ‌‌=sin‌(A−B)
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