Johnny Blaze
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- Aug 8, 2019
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In triangle ABC right angle at A. If Sin B Sin C = 0.2 and Sin (B - C) = 5a/2. Find the value of 5a - 1. Please help me. Thanks.
Here's my working
Since angle A = 90
Angle (B + C) = 90
Sin (B + C) = Sin 90 = 1
Cos (B + C) = Cos 90 = 0
Sin (B + C) = Sin B Cos C + Cos B Sin C
Cos (B + C) = Cos B Cos C - Sin B Sin C
0 = Cos B Cos C - 0.2
Cos B Cos C = 0.2
2 Cos B Cos C = 0.4
Cos (B + C) + Cos (B - C) = 0.4
0 + Cos (B - C) = 0.4
Cos (B - C) = 0.4 = 2/5
Sin (B - C) = Sin B Cos C - Cos B Sin C = 5a/2
Cos (B - C) = Cos B Cos C + Sin B Sin C = 2/5
Now I'm really stuck here. Please help!
[MATH][/MATH]
Here's my working
Since angle A = 90
Angle (B + C) = 90
Sin (B + C) = Sin 90 = 1
Cos (B + C) = Cos 90 = 0
Sin (B + C) = Sin B Cos C + Cos B Sin C
Cos (B + C) = Cos B Cos C - Sin B Sin C
0 = Cos B Cos C - 0.2
Cos B Cos C = 0.2
2 Cos B Cos C = 0.4
Cos (B + C) + Cos (B - C) = 0.4
0 + Cos (B - C) = 0.4
Cos (B - C) = 0.4 = 2/5
Sin (B - C) = Sin B Cos C - Cos B Sin C = 5a/2
Cos (B - C) = Cos B Cos C + Sin B Sin C = 2/5
Now I'm really stuck here. Please help!
[MATH][/MATH]