Question

In a triangle \[ABC\], if \[2s = a + b + c\] and \[\left( {s - b} \right)\left( {s - c} \right) = x\sin^{2}\dfrac{A}{2}\]. Then find the value of \[x\].
A. \[bc\]
B. \[ca\]
C. \[ab\]
D. \[abc\]

Answer 1

Hint: First, we will apply trigonometric functions of half angles of triangles and simplify the expression. Solve the equation by taking the square on both sides to reach the required answer.

Formula used:
Trigonometric functions for half angles: In a triangle \[ABC\]
\[\sin \dfrac{A}{2} = \sqrt {\dfrac{{\left( {s - b} \right)\left( {s - c} \right)}}{{bc}}} \]
\[\sin \dfrac{B}{2} = \sqrt {\dfrac{{\left( {s - a} \right)\left( {s - c} \right)}}{{ac}}} \]
\[\sin \dfrac{C}{2} = \sqrt {\dfrac{{\left( {s - a} \right)\left( {s - b} \right)}}{{ab}}} \]

Complete step by step solution:
The given expressions for a triangle \[ABC\] are:
\[2s = a + b + c\] and \[\left( {s - b} \right)\left( {s - c} \right) = x\sin^{2}\dfrac{A}{2}\]

Let’s apply the half-angle formula for the angle \[A\].
We get,
\[\sin \dfrac{A}{2} = \sqrt {\dfrac{{\left( {s - b} \right)\left( {s - c} \right)}}{{bc}}} \]
Take the square of both sides.
\[\sin^{2} \dfrac{A}{2} = \dfrac{{\left( {s - b} \right)\left( {s - c} \right)}}{{bc}}\]
Multiply both sides by \[bc\].
\[bc \sin^{2} \dfrac{A}{2} = \left( {s - b} \right)\left( {s - c} \right)\]
Substitute the value from the given equation.
\[bc \sin^{2} \dfrac{A}{2} = x\sin^{2}\dfrac{A}{2}\]
Comparing both sides, we get
\[bc = x\]
Hence the correct option is A.

Note: Students do a mistake about the value of \[s\]. \[s\] is the semi-perimeter of the triangle. Thus, the value of \[s\] is \[\dfrac{{a + b + c}}{2}\]. Students should keep in mind that the trigonometry and half-angle formulas are used to determine the exact values of the trigonometric ratios of half of the standard angles. The positive and negative signs are depending on which quadrant the new half-angle is present.