x,y,z are respectively the sines and p,q,r are respectively the cosines of angles A,B,C which are in AP with common difference 2pi/3 i.e 120 degrees then prove that the value of 8x 2 (qy – rz) + 8y 2 (rz – px) + 8z 2 (px – qy) is (27) 1/2

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srikar · Answer

sinA=x sinB=y sinc=z put A = 0 x=o y=sin(120) z=sin(240) 8x 2 (qy – rz)=0 8y 2 (rz – px) = 8sin 2 120(sin240cos240) = 4sin 2 120(sin480) = 4sin 3 120 =4(3*3 1/2 )/8 = 3*3 1/2 /2 8z 2 (px – qy) = 8sin 2 240(-sin120cos120) = 4sin 2 240(-sin240) = 4sin 3 60 = 3*3 1/2 /2 8x 2 (qy – rz) + 8y 2 (rz – px) + 8z 2 (px – qy) = 0 + 3*3 1/2 /2 + 3*3 1/2 /2 = 3*3 1/2 = 27 1/2

Kaustubh Nayyar · Answer

can you please do the ques without taking any of the angle by yourself.

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x,y,z are respectively the sines and p,q,r are respectively the cosines of angles A,B,C which are in AP with common difference 2pi/3 i.e 120 degrees then prove that the value of 8x 2 (qy – rz) + 8y 2 (rz – px) + 8z 2 (px – qy) is (27) 1/2

x,y,z are respectively the sines and p,q,r are respectively the cosines of angles A,B,C which are in AP with common difference 2pi/3 i.e 120 degrees
then prove that the value of
8x²(qy – rz) + 8y²(rz – px) + 8z²(px – qy) is (27)^1/2

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x,y,z are respectively the sines and p,q,r are respectively the cosines of angles A,B,C which are in AP with common difference 2pi/3 i.e 120 degrees then prove that the value of 8x 2 (qy – rz) + 8y 2 (rz – px) + 8z 2 (px – qy) is (27) 1/2

x,y,z are respectively the sines and p,q,r are respectively the cosines of angles A,B,C which are in AP with common difference 2pi/3 i.e 120 degreesthen prove that the value of8x2(qy – rz) + 8y2(rz – px) + 8z2(px – qy) is (27)1/2

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x,y,z are respectively the sines and p,q,r are respectively the cosines of angles A,B,C which are in AP with common difference 2pi/3 i.e 120 degrees
then prove that the value of
8x²(qy – rz) + 8y²(rz – px) + 8z²(px – qy) is (27)^1/2