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Question
If sin A = `3/5` and cos B = `9/41, 0 < "A" < pi/2, 0 < "B" < pi/2`, find the value of cos(A – B)
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Solution
sin A = `3/5`
`0 < "A" < pi/2`
From ΔABC,
AB = `sqrt(5^2 - 3^2)`
= `sqrt(25 - 9)`
= `sqrt(16)`
= 4
cos B = `9/41`
`0 < "B" < pi/2`
From ΔBAD,
AD = `sqrt(41^2 - 9^2)`
= `sqrt((41 + 9)(41 - 9))`
= `sqrt(50 xx 32)`
= `sqrt(100 xx 16)`
= `sqrt(10^2 xx 4^2)`
= 10 × 4
= 40
Now,
From ΔABC, sin A = `3/5`, cos A = `4/5`
From ΔABD, sin B = `40/41`, cos B = `9/41`
cos(A – B) = cos A cos B + sin A sin B
= `(4/5 xx 9/41) + (3/5 xx 40/10)`
= `36/205 + 120/205`
= `156/205`
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