Question

What is the value of x? sin49°=cosx Enter your answer in the box.

Answer 1

Answer:

$x=41^o$

Step-by-step explanation:

We have been given an equation $sin(49^o)=cos(x)$ and we are asked to find the value of x.

Since we know that trigonometric ratios are applied to right triangles and we also know that $sin(a)=cos(b)$, where a and b are the angles other than 90 degree angle.

As one angle of our triangle is given 49 degrees, so to find the value of angle x we need to subtract 49 degrees from 90 degree angle as 49 degree angle and x will be equal to 90 degrees.

$x=90^o-49^o$

$x=41^o$

Therefore, the value of x is 41 degrees.

Answer 2

We know that
if sin A=cos B
then 
angle A and angle B are complementary angles
A+B=90°

in this problem
sin49°=cosx
so
49°+x=90-------> x=90-49--------> x=41°

the answer is
x=41°