Question

If A and B are acute angles and $\sin A = \cos B$ , then find the value of $A + B$ .

Answer 1

Hint : Let us say we have an angle A. Then the value of the sine function of angle A is equal to the value of cosine function of angle $\left( {{{90}^ \circ } - A} \right)$ . Also given that A and B are acute angles which mean each angle measure below 90 degrees. The sum of three angles of a triangle is always 180 degrees. Use this info to further solve the problem.

Complete step-by-step answer :
We are given that A and B are acute angles and $\sin A = \cos B$ .
We have to find the value of $A + B$
A triangle has 3 sides and 3 angles. Three angles of the triangle must sum up to 180 degrees. Which means it can have 3 acute angles or 2 acute angles and one right angle.
Here we are given that A and B are acute angles and $\sin A = \cos B$
We already know that $\sin A$ is also equal to $\cos \left( {{{90}^ \circ } - A} \right)$
Here we have $\sin A = \cos B$ , which means
$\cos \left( {{{90}^ \circ } - A} \right) = \cos B$
Equating the angle values,
${90^ \circ } - A = B \\\ \Rightarrow {90^ \circ } = A + B \\\ \therefore A + B = {90^ \circ } \;$
Therefore the value of $A + B$ is ${90^ \circ }$

Note :: Another approach.
Here we are given that A and B are acute angles and $\sin A = \cos B$
The value of cosine of angle A can also be written as sine of angle $\left( {{{90}^ \circ } - A} \right)$ .
Here we have $\sin A = \cos B$ , convert cosine function into sine function by using the above conversion.
This results $\cos B = \sin \left( {{{90}^ \circ } - B} \right)$
On substituting the above value in $\sin A = \cos B$ , we get
$\sin A = \cos B \\\ \Rightarrow \sin A = \sin \left( {{{90}^ \circ } - B} \right) \\\ \Rightarrow A = {90^ \circ } - B \\\ \therefore A + B = {90^ \circ } \;$
Therefore the value of $A + B$ is ${90^ \circ }$ , which means the remaining angle C of the triangle measures ${90^ \circ }$ which is a right angle. So the triangle has 2 acute angles and one right angle.