Question

If $\vartriangle ABC$ is right angled at \[C\], then what is the value of \[cos\left( {A + B} \right)\]?
A. $0$
B. $1$
C. $\dfrac{1}{2}$
D. $\dfrac{{\sqrt 3 }}{2}$

Answer 1

Hint: From the question we have to find the value of required trigonometry relation. The given of the question is a triangle ABC with right angled at C. By using the angle sum property of a triangle we will get the value of addition of the required two angles. On substituting the value into the trigonometric relation we can get the required solution.
Angle sum property of a triangle states that the sum of all the angles of a triangle is $180^\circ $.

Complete step-by-step answer:

It is given that $\angle C$ is right angled in a $\vartriangle ABC$.
We already know that, angle sum property of a triangle states that the sum of all the angles of a triangle is $180^\circ $. By using the angle sum property of a triangle for $\vartriangle ABC$, we get,
$\angle A + \angle B + \angle C = 180^\circ $ (Angle sum property of a triangle)
And in the given of the question, where $\angle C = 90^\circ $
Substituting the value of \[C\] in the above equation and solving,
$\Rightarrow$$\angle A + \angle B + 90^\circ = 180^\circ $
Rearranging the terms to solve,
$\Rightarrow$$\angle A + \angle B = 180^\circ - 90^\circ $
Subtracting the term,
Hence,
$\Rightarrow$$\angle A + \angle B = 90^\circ $
Now we get the addition value of the $\angle A$ and $\angle B$
Now, we have to find the value of \[cos\left( {A + B} \right)\]
We will substitute this in \[cos\left( {A + B} \right)\]
$\Rightarrow$$ \Rightarrow \cos (A + B)$
As we have derived that $\angle A + \angle B = 90^\circ $, substituting this and by using $\cos (90^\circ ) = 0$ we get,
$ \Rightarrow \cos (90^\circ ) = 0$

The value of \[cos\left( {A + B} \right)\] is equal to 0.

Note: The vertex at which the triangle is making a right angle must be carefully observed as it is very important and can change the entire question. For example, if the triangle had been right angled at $A$, $\cos (B + C) = 90^\circ $. In this case, the answer would not have been the same.