Question

How do you find tan2A, given sin A = 3/5 and A is in QII?

Answer 1

ANSWER :

EXPLANATION:

sin A=3/5

sinA=opposite side/hypotenuse

then Tan A=opposite side/Adjacent side

so. Tan2A=?

by Pythagoras theorem_.

(hyp)2=(side)2 + (side)2

(5)2=(3)2. + (x)2

25= 9+(x)2

25 - 9 = (x)2

16 = (x)2

√16 = x

4=x

then

tan2A = (3/4)2

(3)2/(4)2

9/16

there fore the sinA = 3/5

then Tan2A =9/16