5sin A - 2cos A(b), If 13 sin A =5 and A is acute then find the value

1.	5sin A - 2cos A(b), If 13 sin A =5 and A is acute then find the value oftan A
Answer» sin A = 5/13cos A = (1 - sin^2 A)^1/2 = (1 - 25/169)^1/2 = [(169 - 25)/169]^1/2 = (144/169)^1/2 = 12/13tan A = sin A/cos A = 5/13 / 12/13 = 5/12 Therefore,Value of 5sin A - 2cos A/tan A= (55/13 - 212/13)/(5/12)= (25/13 - 24/13)/(5/12)= (1/13)/(5/12)= 12/65

5sin A - 2cos A(b), If 13 sin A =5 and A is acute then find the value oftan A

Answer»

sin A = 5/13cos A = (1 - sin^2 A)^1/2 = (1 - 25/169)^1/2 = [(169 - 25)/169]^1/2 = (144/169)^1/2 = 12/13tan A = sin A/cos A = 5/13 / 12/13 = 5/12

Therefore,Value of 5sin A - 2cos A/tan A= (5*5/13 - 2*12/13)/(5/12)= (25/13 - 24/13)/(5/12)= (1/13)/(5/12)= 12/65