Question

In a $∆ABC$, if the sides are $a=3,b=5$ and $c=4$, then $\sin \left(\frac{B}{2}\right)+\cos \left(\frac{B}{2}\right)$ is equal to

• A. $\sqrt{2}$
• B. $\frac{\sqrt{3+1}}{2}$
• C. $\frac{\sqrt{3-1}}{2}$
• D. $1$

Answer 1

The correct option is A

$\sqrt{2}$

Explanation for correct option:

Step 1: Finding squares of side and checking for Pythagoras theorem:

Given a triangle $ABC$ in which sides are $a=3,b=5$ and $c=4$

Now applying Pythagoras theorem we get,

$$\begin{gathered} b^2=a^2+c^2 \\ \text{LHS}: \\ =5^2 \\ =25 \\ \text{RHS:} \\ =3^2+4^2 \\ =9+16 \\ =25 \end{gathered}$$

Since, LHS $=$RHS, the triangle $ABC$ is a right-angled triangle at $B$.

Hence, $\angle B=90°$

Step 2: Finding the value of the expression $\sin \left(\frac{B}{2}\right)+\cos \left(\frac{B}{2}\right)$:

$$\begin{gathered} \therefore \sin \left(\frac{B}{2}\right)+\cos \left(\frac{B}{2}\right)=\sin {\left(45\right)}^0+\cos {\left(45\right)}^0 \\ =\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}} \\ =\frac{2}{\sqrt{2}} \\ =\sqrt{2} \end{gathered}$$

Therefore, the value of the given expression is equal to $\sqrt{2}$.

Hence, the correct answer is option (A).