Given triangle JKL, sin 38 degree equals

Question

asked Jun 12, 2018 165k views

2 Answers

Miran · Answer 1 · 2018-06-18T12:35:00+0000

Answer:

B.
$\text{cos}(52^(\circ))$ .

Explanation:

We have been given a right triangle. We are asked to complete our given statement.

$\text{sin}(38^(\circ))$ equals ___.

We will use identity
$\text{sin}(x)=\text{cos}(90-x)$ to solve our given problem.

$\text{sin}(38^(\circ))=\text{cos}(90^(\circ)-38^(\circ))$

$\text{sin}(38^(\circ))=\text{cos}(52^(\circ))$

Therefore,
$\text{sin}(38^(\circ))$ equals
$\text{cos}(52^(\circ))$ .