Given that sin x = 3/5, 0 ≤ x ≤ 90, evaluate (tanx + 2cosx)
2\(\frac{11}{20}\)
\(\frac{11}{20}\)
2\(\frac{7}{20}\)
\(\frac{1}{20}\)
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 \(\frac{7}{20}\)