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Question
If `sin"A"/3 = sin"B"/4 = 1/5` and A, B are angles in the second quadrant then prove that 4cosA + 3cosB = – 5.
Solution
Given, `sin"A"/3 = sin"B"/4 = 1/5`
∴ sin A = `3/5 and sin "B" = 4/5`
We know that,
cos2A = 1 – sin2A
= `1 - (3/5)^2`
= `1 - 9/25`
∴ cos2A = `16/25`
∴ cos A = `± 4/5` ...[Taking square root on both the sides]
Since A lies in the second quadrant,
cos A < 0
∴ cos A = `- 4/5`
sin B = `4/5`
We know that,
cos2B = 1 – sin2B
= `1 - (4/5)^2`
= `1 - 16/25`
cos2B = `9/25`
∴ cos B = `± 3/5` ...[Taking square root on both the sides]
Since B lies in the second quadrant
cos B < 0
∴ cos B = `-3/5`
∴ 4cos A + 3cos B = `4(-4/5) + 3(-3/5)`
= `-16/5 - 9/5`
= `-25/5`
= – 5
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