Question

If PQ and PR are two sides of a triangle, then the angle between them which gives maximum area of the triangle is:
$\left( 1 \right){\pi}$
$\left( 2 \right)\dfrac{\pi }{3}$
$\left( 3 \right)\dfrac{\pi }{4}$
$\left( 4 \right)\dfrac{\pi }{2}$

Answer 1

Hint: To solve this question, we will use the formula of the area of the triangle that is $Area = \dfrac{1}{2} \times PR \times PQ \times \sin \alpha $. Then, we will use the condition that the maximum value of sine is $1$. So, we will equate the angle of the formula to $1$ and will simplify it to find the answer.

Complete step-by-step solution:
Since, it is given that PQ and PR are two sides of a triangle.
Here, let the triangle is PQR and the angle between the sides PR and PQ is $\alpha $ as,

Now, the area of a triangle can be obtained by the half of the product of two sides and the sine value of angle between then as,
$Area = \dfrac{1}{2} \times PR \times PQ \times \sin \alpha $
As we know that the value of sine lies between $ - 1$ and $1$. Since, the maximum value of sine is $1$. So, The maximum area will be obtained when the sine value of angle between both sides is $1$.
$Area = \dfrac{1}{2} \times PR \times PQ$
Now, we will calculate the value of $\alpha $ as,
$ \Rightarrow \sin \alpha = 1$
Here, we will substitute $\sin \dfrac{\pi }{2}$ for $1$ in the above step because the value of sine is $1$ at $\dfrac{\pi }{2}$ .
$ \Rightarrow \sin \alpha = \sin \dfrac{\pi }{2}$
After cancelling out sine, we will have the value of $\alpha $:
$ \Rightarrow \alpha = \dfrac{\pi }{2}$
Hence, option D is the right option.

Note: Here is the table of value of sine for different angles as,

Angle	$0$	$\dfrac{\pi }{6}$	$\dfrac{\pi }{4}$	$\dfrac{\pi }{3}$	$\dfrac{\pi }{2}$
Sine	$0$	$\dfrac{1}{2}$	$\dfrac{1}{{\sqrt 2 }}$	$\dfrac{{\sqrt 3 }}{2}$	$1$

Since, the negative value of the sine does not satisfy the area of the triangle because the area never be in negative and in the positive values, we can clearly observe from the table that all the values have multiplication of half also except $0$ and $1$. As we know that $1$ is greater than $0$. Hence, the sine value would be $1$ for getting the maximum area of a triangle.