Question

 In a triangle \[ABC\] , find the value of \[\dfrac{{\sin B}}{{\sin\left( {A + B} \right)}}\] .
A. \[\dfrac{b}{{a + b}}\]
B. \[\dfrac{b}{c}\]
C. \[\dfrac{c}{b}\]
D. None of these

Answer 1

Hint:
First, simplify the denominator of the required equation by using the sum of angles of a triangle and trigonometric formula of the sine angle. Then solve it by using the sine law to get the required answer.



Formula Used:
The sine law: For any triangle \[ABC\], \[\dfrac{{\sin A}}{a} = \dfrac{{\sin B}}{b} = \dfrac{{\sin C}}{c}\]
\[\sin\left( {\pi - \theta } \right) = \sin \theta \]



Complete step-by-step answer:
In \[\triangle ABC \], using angle sum property of triangle
\[A + B + C = \pi \]
\[ \Rightarrow A + B = \pi - C\]

Let’s simplify the given equation.
\[\dfrac{{\sin B}}{{\sin\left( {A + B} \right)}} = \dfrac{{\sin B}}{{\sin\left( {\pi - C} \right)}}\]
Apply the trigonometric identity \[\sin\left( {\pi - \theta } \right) = \sin \theta \]
\[\dfrac{{\sin B}}{{\sin\left( {A + B} \right)}} = \dfrac{{\sin B}}{{\sin C}}\]
Now apply the sine law \[\dfrac{{\sin B}}{b} = \dfrac{{\sin C}}{c}\] on the right-hand side.
We get,
\[\dfrac{{\sin B}}{{\sin\left( {A + B} \right)}} = \dfrac{b}{c}\]
Hence the correct option is B.



Note:
The law of sine or the sine law states that the ratio of the side length of a triangle to the sine of the opposite angle, is the same for all three sides.
The sum of the internal angles of the triangle is \[180^ {\circ }\]. It is also denoted by \[\pi \].