Dalam segitiga ABC dik sin A-sin B=setengah akar(2+akar 3) dan cos A - cos B = setengah akar(2-akar3) maka A-B= brp drajat?
Galladeaviero
. (sin A - sin B)² = 1/4 (2 + √3) sin² A +sin² B - 2 sin A sin B = 1/2 ( 2+√3) ....(1) . (cos A - cos B)² = 1/4(2 - √3) cos² A+ cos² B - 2 cos A cos B = 1/2 (2 -√3)....(2). .. (1) + (2) = 1+1 - 2 (sin A sin B + cos A cos B) = 1/2 { 2+√3 +2 -√3) 2 - 2 { cos (A - B)} = 1/2(4) = 2 - 2 cos (A - B) = 2 -2 = 0 cos (A - B) = 0 = cos 90 (A-B)= 90°
(sin A - sin B)² = 1/4 (2 + √3)
sin² A +sin² B - 2 sin A sin B = 1/2 ( 2+√3) ....(1)
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(cos A - cos B)² = 1/4(2 - √3)
cos² A+ cos² B - 2 cos A cos B = 1/2 (2 -√3)....(2).
..
(1) + (2) =
1+1 - 2 (sin A sin B + cos A cos B) = 1/2 { 2+√3 +2 -√3)
2 - 2 { cos (A - B)} = 1/2(4) = 2
- 2 cos (A - B) = 2 -2 = 0
cos (A - B) = 0 = cos 90
(A-B)= 90°