Question

3) In ΔABC ef sin A = cosB then angle C= ? .
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Answer 1

Step-by-step explanation:

 sinA−cosB=cosC

⇒sinA=cosB−cosC

⇒2sin2Acos2A=2sin2Acos(2B−C)

⇒cos2A=cos(2B−C)[∵sin2A=0]

⇒2A=2B−C⇒A=B−C

But A+B+C=π

Therefore, 2B=π

⇒B=2π

angel c=90°

Answer 2

Answer:

answer is simple

angle c= 90⁰

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