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A,B,C and D are the angles of a cyclic quadrilateral. Prove that:cosA+cosB+cosc+cosD=0 |
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Answer» As A,B,C,D are the angles of a cyclic quadrilateral.so, ANGLE A + angle C= 180° = >angle A =180°-C Cos A = -CosC angle C= 180°- A Cos C= - Cos A SILLY angle B +angle D= 180° angle B=180°- D Cos B= -Cos D angle D=180° -B Cos D= -Cos B Now,L:H:S : CosA + CosB + Cos C+ Cos D = Cos A + Cos B + Cos ( 180°- A)+ Cos (180°- B)= Cos A+Cos B-Cos A -Cos B= 0(R:H:S)please mark it as BRAINLIST if it HELPS ✨ |
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