Question

If in a triangle ABC, \[\left( {s - a} \right)\left( {s - b} \right) = s\left( {s - c} \right)\], then find angle C.
A. \[{90^ \circ }\]
B. \[{45^ \circ }\]
C. \[{30^ \circ }\]
D. \[{60^ \circ }\]

Answer 1

Hint: To solve this question we will apply the half-angle formula of a triangle. To calculate angle C, we will use half angle formula of \[\sin \dfrac{C}{2}\] and \[\cos \dfrac{C}{2}\]. Then we will calculate \[\tan \dfrac{C}{2}\]and by solving this we can calculate angle C.

Formula used:
Half angle formula for triangle:
\[\sin \dfrac{C}{2} = \sqrt {\dfrac{{\left( {s - a} \right)\left( {s - b} \right)}}{{ab}}} \]
\[\cos \dfrac{C}{2} = \sqrt {\dfrac{{s\left( {s - c} \right)}}{{ab}}} \]
Trigonometry formula
\[\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}\]

Complete step by step solution:
Given equation is \[\left( {s - a} \right)\left( {s - b} \right) = s\left( {s - c} \right)\]
Rewrite the above equation:
\[\dfrac{{\left( {s - a} \right)\left( {s - b} \right)}}{{s\left( {s - c} \right)}} = 1\] ……(i)
We know the half-angle formulas that are
\[\sin \dfrac{C}{2} = \sqrt {\dfrac{{\left( {s - a} \right)\left( {s - b} \right)}}{{ab}}} \] …..(ii)
And \[\cos \dfrac{C}{2} = \sqrt {\dfrac{{s\left( {s - c} \right)}}{{ab}}} \] …..(iii)
Divide (ii) by (iii)
\[\dfrac{{\sin \dfrac{C}{2}}}{{\cos \dfrac{C}{2}}} = \dfrac{{\sqrt {\dfrac{{\left( {s - a} \right)\left( {s - b} \right)}}{{ab}}} }}{{\sqrt {\dfrac{{s\left( {s - c} \right)}}{{ab}}} }}\]
Apply the formula \[\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}\]
\[\tan \dfrac{C}{2} = \dfrac{{\sqrt {\dfrac{{\left( {s - a} \right)\left( {s - b} \right)}}{{ab}}} }}{{\sqrt {\dfrac{{s\left( {s - c} \right)}}{{ab}}} }}\]
Simplify the right side of the equation:
\[\tan \dfrac{C}{2} = \sqrt {\dfrac{{\left( {s - a} \right)\left( {s - b} \right)}}{{ab}}} \times \sqrt {\dfrac{{ab}}{{s\left( {s - c} \right)}}} \]
Cancel out \[\sqrt {ab} \]
\[\tan \dfrac{C}{2} = \sqrt {\dfrac{{\left( {s - a} \right)\left( {s - b} \right)}}{{s\left( {s - c} \right)}}} \]
From equation (i) we get \[\dfrac{{\left( {s - a} \right)\left( {s - b} \right)}}{{s\left( {s - c} \right)}} = 1\].
Substitute \[\dfrac{{\left( {s - a} \right)\left( {s - b} \right)}}{{s\left( {s - c} \right)}} = 1\] in the above equation:
\[\tan \dfrac{C}{2} = \sqrt 1 \]
\[ \Rightarrow \tan \dfrac{C}{2} = 1\]
We know that, \[\tan \dfrac{\pi }{4} = 1\]
\[ \Rightarrow \tan \dfrac{C}{2} = \tan \dfrac{\pi }{4}\]
Compare both sides of the above equation:
\[ \Rightarrow \dfrac{C}{2} = \dfrac{\pi }{4}\]
Multiply both sides by 2
\[ \Rightarrow C = \dfrac{\pi }{4} \times 2\]
\[ \Rightarrow C = \dfrac{\pi }{2}\]
\[ \Rightarrow C = {90^ \circ }\]
Hence option A is the correct option.

Note: We can solve the given question by using the given equation. Divide the denominator and numerator by ab of the equation \[\dfrac{{\left( {s - a} \right)\left( {s - b} \right)}}{{s\left( {s - c} \right)}} = 1\]. The equation becomes \[\dfrac{{\dfrac{{\left( {s - a} \right)\left( {s - b} \right)}}{{ab}}}}{{\dfrac{{s\left( {s - c} \right)}}{{ab}}}} = 1\]. Then using half angle formula, we get \[\dfrac{{{{\sin }^2}\dfrac{C}{2}}}{{{{\cos }^2}\dfrac{C}{2}}} = 1\]. By solving the equation, we can get the measurement of angle C.