Question

 In \[\Delta ABC\], if the sides are \[a = 3,b = 5\] and \[c = 4\] then find the value of \[\sin \left( {\dfrac{B}{2}} \right) + \sin \left( {\dfrac{B}{2}} \right)\].
A. \[\sqrt 2 \]
B. \[\dfrac{{\sqrt {3 + 1} }}{2}\]
C. \[\dfrac{{\sqrt {3 - 1} }}{2}\]
D. \[1\]

Answer 1

Hint:
In the given question, we need to find the value of \[\sin \left( {\dfrac{B}{2}} \right) + \sin \left( {\dfrac{B}{2}} \right)\]. For this, we will use Pythagoras theorem to get the angle \[B\]. Using this angle \[B\] , we will get the desired result.

Formula used:
Let the sides of the right angled triangle \[ABC\] be \[a, b, and c\].
By pythagoras theorem: \[{b^2} = {a^2} + {c^2}\]



Complete step-by-step answer:
Consider the following figure.

Now, we will apply Pythagoras theorem in a triangle \[ABC\].
\[ \Rightarrow {b^2} = {a^2} + {c^2}\]
Given, \[a = 3,b = 5\] and \[c = 4\]
\[\Rightarrow {5^2} = {3^2} + {4^2}\]
\[\Rightarrow 25 = 9 + 16\]
\[\Rightarrow 25 = 25\]
Therefore, the triangle \[ABC\] is a right angled triangle at angle \[B\].
Thus, we get \[\angle B = {90^o}\].

Now, we will find the value of \[\sin \left( {\dfrac{B}{2}} \right) + \sin \left( {\dfrac{B}{2}} \right)\].
\[ \Rightarrow \sin \left( {\dfrac{B}{2}} \right) + \sin \left( {\dfrac{B}{2}} \right) = \sin \left( {\dfrac{{90}}{2}} \right) + \sin \left( {\dfrac{{90}}{2}} \right)\]
\[ \Rightarrow \sin \left( {\dfrac{B}{2}} \right) + \sin \left( {\dfrac{B}{2}} \right) = \sin \left( {45} \right) + \sin \left( {45} \right)\]
But \[\sin ({45^o}) = \dfrac{1}{{\sqrt 2 }}\] and \[\cos ({45^o}) = \dfrac{1}{{\sqrt 2 }}\]
\[\Rightarrow \sin \left( {\dfrac{B}{2}} \right) + \sin \left( {\dfrac{B}{2}} \right) = \dfrac{1}{{\sqrt 2 }} + \dfrac{1}{{\sqrt 2 }}\]
\[\Rightarrow \sin \left( {\dfrac{B}{2}} \right) + \sin \left( {\dfrac{B}{2}} \right) = \sqrt 2 \]
Hence, the value of \[\sin \left( {\dfrac{B}{2}} \right) + \sin \left( {\dfrac{B}{2}} \right)\] is \[\sqrt 2 \].

Therefore, the correct option is (A).


Note: Many students make mistakes in the calculation part as well as writing Pythagoras theorem. This is the way through which we can solve the example in the simplest way. Also, it is essential to get the correct value of \[\angle B\] to get the desired result.