Question

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

Answer 1

`x=41^o`

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°