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JEE Advance - Mathematics (2023 - Paper 2 Online - No. 15)

Consider an obtuse angled triangle $A B C$ in which the difference between the largest and the smallest angle is $\frac{\pi}{2}$ and whose sides are in arithmetic progression. Suppose that the vertices of this triangle lie on a circle of radius 1.
Consider an obtuse angled triangle $A B C$ in which the difference between the largest and the smallest angle is $\frac{\pi}{2}$ and whose sides are in arithmetic progression. Suppose that the vertices of this triangle lie on a circle of radius 1.
Consider an obtuse angled triangle $A B C$ in which the difference between the largest and the smallest angle is $\frac{\pi}{2}$ and whose sides are in arithmetic progression. Suppose that the vertices of this triangle lie on a circle of radius 1.
$$ \text { Then the inradius of the triangle } A B C \text { is } $$ :
Answer
0.25

Explanation

In radius $ r=\frac{\Delta}{S}=\left[\frac{a}{2 R(\sin A+\sin B+\sin C)}\right)$

$$ \begin{aligned} r & =\frac{a}{\sin \left(\frac{\pi}{2}-2 C\right) \sin \left(\frac{\pi}{2}+C\right)+\sin C} \\\\ & =\frac{a}{\cos 2 C+\cos C+\sin C} \\\\ & =\frac{a}{\cos 2 C+\sqrt{1+\sin 2 C}} \\\\ & =\frac{\frac{3 \sqrt{7}}{16}}{\sqrt{\frac{7}{4}}+\sqrt{\frac{7}{2}}}=\frac{1}{4} \\\\ \Rightarrow r & =\frac{1}{4}=0.25 \\\\ \Rightarrow r & =0.25 \end{aligned} $$

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