Question
In ∆ABC if cosA = sin B – cos C then show that it is a right angled triangle.
$\therefore 2 \cos \left(\frac{A+C}{2}\right) \cdot \cos \left(\frac{A-C}{2}\right)=\sin B$
$\therefore 2 \cos \left(\frac{\pi}{2}-\frac{B}{2}\right) \cdot \cos \left(\frac{A-C}{2}\right)=\sin B$
$\ldots[\because \mathrm{A}+\mathrm{B}+\mathrm{C}=\pi]$
$\therefore 2 \sin \frac{B}{2} \cdot \cos \left(\frac{A-C}{2}\right)=2 \sin \frac{B}{2} \cdot \cos \frac{B}{2}$
$\therefore \cos \left(\frac{A-C}{2}\right)=\cos \frac{B}{2}$
$\therefore \frac{A-C}{2}=\frac{B}{2}$
∴ A – C = B ∴ A = B + C ∴ A + B + C = 180° gives A + A = 180° ∴ 2A = 180 ∴ A = 90° ∴ ∆ ABC is a rightangled triangle.
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unit from the base point A(1, 2, 3).