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標題:
Maths problem
發問:
1) prove the identites. cos^2(90-a)-sin^2(90-a)=1-2cos^2a 2)Find the value of x if sin(x+50)+sin(40-x)tan(x+50)=√3 3)In triangle ABC,AC=BC=2 and angleACB=45degree. Find the area of triangle ABC. Picture:http://postimg.org/image/au4yjog45/ NEED STEP,PLZ!!!
最佳解答:
(1) cos^2 (90 - a) - sin^2(90 - a) = sin^a - cos^a = (1 - cos^2a) - cos^2a = 1 - 2 cos^2 a. (2) sin ( x + 50) + sin(40 - x)tan(x + 50) = sqrt 3 sin[90 - (40 - x)] + sin ( 40 - x) tan [ 90 - (40 - x)] = sqrt 3 cos(40 - x) + sin(40 - x)/tan ( 40 - x) = sqrt 3 cos(40 - x) + cos(40 - x) = sqrt 3 cos(40 - x) = sqrt 3/2 40 - x = arc cos(sqrt3/2) = 30 so x = 40 - 30 = 10 degree. (3) X is a point on AC such that BX is perpendicular to AC. BX/BC = sin 45 BX = BC sin 45 = 2 x sqrt 2/2 = sqrt 2. so area of triangle = (AC x BX)/2 = (2 sqrt 2)/2 = sqrt 2.
還是要去 http://aaashops。com 品質不錯，老婆很喜歡。 哲住厭亣兣
Maths problem
發問:
1) prove the identites. cos^2(90-a)-sin^2(90-a)=1-2cos^2a 2)Find the value of x if sin(x+50)+sin(40-x)tan(x+50)=√3 3)In triangle ABC,AC=BC=2 and angleACB=45degree. Find the area of triangle ABC. Picture:http://postimg.org/image/au4yjog45/ NEED STEP,PLZ!!!
最佳解答:
(1) cos^2 (90 - a) - sin^2(90 - a) = sin^a - cos^a = (1 - cos^2a) - cos^2a = 1 - 2 cos^2 a. (2) sin ( x + 50) + sin(40 - x)tan(x + 50) = sqrt 3 sin[90 - (40 - x)] + sin ( 40 - x) tan [ 90 - (40 - x)] = sqrt 3 cos(40 - x) + sin(40 - x)/tan ( 40 - x) = sqrt 3 cos(40 - x) + cos(40 - x) = sqrt 3 cos(40 - x) = sqrt 3/2 40 - x = arc cos(sqrt3/2) = 30 so x = 40 - 30 = 10 degree. (3) X is a point on AC such that BX is perpendicular to AC. BX/BC = sin 45 BX = BC sin 45 = 2 x sqrt 2/2 = sqrt 2. so area of triangle = (AC x BX)/2 = (2 sqrt 2)/2 = sqrt 2.
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其他解答:還是要去 http://aaashops。com 品質不錯，老婆很喜歡。 哲住厭亣兣
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