Question

If 0° < A, B < 90° such that cos A = $\frac{5}{13}$ and sin B = $\frac{4}{5}$, find the value of sin (A - B).

Answer 1

given, If 0° < A, B < 90° such that cos A = $\frac{5}{13}$ and sin B = $\frac{4}{5}$,

cosA = 5/13 = b/h

so, b = 5 and h = 13

then, p = √(h² - b²) = √(13² - 5²) = 12

so, sinA = p/h = 12/13

similarly, sinB = 4/5 = p/h

so, p = 4 and h = 5

then, b = √(h² - p²) = √(5² - 4²) = 3

so, cosB = b/h = 3/5


now, sin(A - B) = sinA. cosB - cosA. sinB [ from formula ]

= 12/13 × 3/5 - 5/13 × 4/5

= 36/65 - 20/65

= 16/65

hence, sin(A - B) = 16/65

Answer 2

HELLO DEAR,




given:- 0° < A, B < 90° such that cos A = $\frac{5}{13}$ and sin B = $\frac{4}{5}$

cosA = 5/13 = b/h

so, b = 5 and h = 13

then, p = √(h² - b²) = √(13² - 5²) = 12

so, sinA = p/h = 12/13

similarly, sinB = 4/5 = p/h

so, p = 4 and h = 5

then, b = √(h² - p²) = √(5² - 4²) = 3

so, cosB = b/h = 3/5


now, sin(A - B) = sinA. cosB - cosA. sinB

=> 12/13 × 3/5 - 5/13 × 4/5

=> 36/65 - 20/65

=> 16/65

hence, sin(A - B) = 16/65



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THANKS