if triangle ABC is right angled at C, then the value of sin(A+B) =? 10th
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since ABC is right angled and angle C is 90°
therefore,
A+B=180°-C
A+B=180°-90°
A+B= 90°
Therefore,cos (A+B)=cos90°
=0
therefore,
A+B=180°-C
A+B=180°-90°
A+B= 90°
Therefore,cos (A+B)=cos90°
=0
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