Given that sin x = 3/5, 0 ≤ x ≤ 90, evaluate (tanx + 2cosx)

A.

2\(\frac{11}{20}\)

B.

\(\frac{11}{20}\)

C.

2\(\frac{7}{20}\)

D.

\(\frac{1}{20}\)

Correct answer is B

Sin x = \(\frac{opp}{hyp}\)

sinx = \(\frac{3}{5}\)

using Pythagoras' theorem

 hyp\(^2\) = opp\(^2\) + adj\(^2\)

adj\(^2\) = 5\(^2\) - 3\(^2\) = 25 - 9

adj\(^2\) = 16 

adj =  16 

adj = 4.

tanx = \(\frac{opp}{adj}\)

= \(\frac{3}{4}\)

cosx = \(\frac{adj}{hyp}\)

= \(\frac{4}{5}\)

(tanx + 2cosx) = \(\frac{3}{4}\) + 2(\(\frac{4}{5}\))

= \(\frac{15 + 32}{20}\)

= \(\frac{47}{20}\) or 

2 \(\frac{7}{20}\)