JEE Advanced 2023 · Paper 2 · Mathematics Q15

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MathematicsNumerical+3Question 15 of 51

Passage

PARAGRAPH "I"

Consider an obtuse angled triangle $ABC$ in which the difference between the largest and the smallest angle is $\dfrac\pi2$ and whose sides are in arithmetic progression. Suppose that the vertices of this triangle lie on a circle of radius 1.

Then the inradius of the triangle $ABC$ is _____.

Your answer

Answer and worked solution

Official answer: 0.24

From the previous part: the area is $\Delta=\frac{3\sqrt7}{16}$ and $\cos2C=\frac{\sqrt7}4$ .

Semi-perimeter. The sides are in AP, so $s=\frac{3b}2$ , where
$b=2R\sin B=2\cos2C=\frac{\sqrt7}2.$
So $s=\frac{3\sqrt7}4$ .

Inradius.
$r=\frac\Delta s=\frac{3\sqrt7/16}{3\sqrt7/4}=\frac14=0.25.$

Topics:Properties of triangles Trigonometric ratios and identities

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JEE Advanced 2023 · Paper 2, Mathematics question 15