Consider an obtuse-angled triangle ABC in whi | vedclass

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Question

(A) Let the angles of the triangle be $A, C, B$ such that $A < C < B$. Given $B - A = \frac{\pi}{2}$.Let the sides be $a_1 = n-d, a_2 = n, a_3 = n+d$

Vedclass · Accepted Answer

$1008, 0.25$

Vedclass · Answer

(A) Let the angles of the triangle be $A, C, B$ such that $A Let the sides be $a_1 = n-d, a_2 = n, a_3 = n+d$ in arithmetic progression.Using the sine rule with circumradius $R=1$,the sides are $2R \sin A, 2R \sin C, 2R \sin B$.Thus,$n-d = 2 \sin A$,$n = 2 \sin C$,$n+d = 2 \sin B$.Since $A+B+C = \pi$ and $B = A + \frac{\pi}{2}$,we have $C = \pi - (A + B) = \pi - (2A + \frac{\pi}{2}) = \frac{\pi}{2} - 2A$.Since $A+C+B = \pi$,$A + (\frac{\pi}{2} - 2A) + (A + \frac{\pi}{2}) = \pi$,which is consistent.From $2n = (n-d) + (n+d) = 2 \sin A + 2 \sin B$,we get $n = \sin A + \sin B = \sin A + \cos A$.Also $n = 2 \sin C = 2 \sin(\frac{\pi}{2} - 2A) = 2 \cos 2A$.Equating $n$: $\sin A + \cos A = 2 \cos 2A = 2(\cos^2 A - \sin^2 A) = 2(\cos A - \sin A)(\cos A + \sin A)$.Since $\sin A + \cos A 
eq 0$,we have $1 = 2(\cos A - \sin A) \Rightarrow \cos A - \sin A = \frac{1}{2}$.Squaring both sides: $1 - 2 \sin A \cos A = \frac{1}{4} \Rightarrow \sin 2A = \frac{3}{4}$.Area $a = \frac{1}{2} (n-d)(n+d) \sin C = \frac{1}{2} (2 \sin A)(2 \sin B) \sin C = 2 \sin A \cos A \sin C = \sin 2A \cos 2A$.Since $\sin 2A = \frac{3}{4}$,$\cos 2A = \sqrt{1 - (3/4)^2} = \frac{\sqrt{7}}{4}$.$a = \frac{3}{4} 	imes \frac{\sqrt{7}}{4} = \frac{3\sqrt{7}}{16}$.$(64a)^2 = (64 	imes \frac{3\sqrt{7}}{16})^2 = (4 	imes 3\sqrt{7})^2 = 144 	imes 7 = 1008$.Inradius $r = \frac{a}{s} = \frac{a}{(3n/2)} = \frac{2a}{3n} = \frac{2 \sin 2A \cos 2A}{3(2 \cos 2A)} = \frac{\sin 2A}{3} = \frac{3/4}{3} = \frac{1}{4} = 0.25$.

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