If A and B are acute angles and $ \sin A = \cos B $ , then find the value of $ A + B $ .
Answer
602.7k+ views
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.
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.
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