Question

if 13sin A=5 and A is acute find the value of

Answer 1

see the pic for answer

Answer 2

Given:

13 sinA = 5

sinA = 5/13 = BC/AC
( perpendicular/hypotenuse)

In the figure, PERPENDICULAR(BC)= 5 & HYPOTENUSE(AC)= 13

By Pythagoras theorem
AC² = AB²+BC²

13²= AB²+ 5²

169 = AB² + 25

169 - 25 = AB²

144 = AB²

AB =√ 144

AB = 12

Find
cosA = AB/AC (B/H = 12/13

tanA = BC/AB (P/B) = 5/12

(5 sinA - 2cosA)/tanA

= (5 × 5/13)- (2×12/13) / 5/12

= (25/13 - 24/13) /5/12

= (1/13) / 5/12

= (1/13) × (12/5)

=12/65


(5 sinA - 2cosA)/tanA= 12/65

Hence, the value of
(5 sinA - 2cosA)/tanA = 12/65

==================================================================
Hope this will help you...