MHT CET · Maths · Properties of Triangles
In \(\triangle \mathrm{ABC}\), with usual notations, \(2 \mathrm{ac} \sin \left(\frac{1}{2}(\mathrm{~A}-\mathrm{B}+\mathrm{C})\right)\) is equal to
- A \(a^2+b^2-c^2\)
- B \(c^2+a^2-b^2\)
- C \(b^2-c^2-a^2\)
- D \(c^2-a^2-b^2\)
Answer & Solution
Correct Answer
(B) \(c^2+a^2-b^2\)
Step-by-step Solution
Detailed explanation
\(\begin{aligned} 2 \mathrm{ac} \sin \frac{\mathrm{A}-\mathrm{B}+\mathrm{C}}{2} & =2 \mathrm{ac} \sin \frac{\pi-2 \mathrm{~B}}{2} \\ & =2 \mathrm{ac} \cos \mathrm{B}\end{aligned}\)
\(
=2 \mathrm{ac} \frac{\mathrm{c}^2+\mathrm{a}^2-\mathrm{b}^2}{2 \mathrm{ca}}
\)....[By cosine rule \(]\)
\(=c^2+a^2-b^2\)
\(
=2 \mathrm{ac} \frac{\mathrm{c}^2+\mathrm{a}^2-\mathrm{b}^2}{2 \mathrm{ca}}
\)....[By cosine rule \(]\)
\(=c^2+a^2-b^2\)
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