JEE Advanced · 2023 · 04 Jun 2023 · Paper 2 · Properties of Triangle

Passage

Consider an obtuse angled triangle ABC in which the difference between the largest and the smallest angle is

\frac{π}{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 ABC in which the difference between the largest and the smallest angle is

\frac{π}{2}

and whose sides are in arithmetic progression. Suppose that the vertices of this triangle lie on a circle of radius 1.

Let a be the area of the triangle ABC. Then the value of

(64 a)^{2}

Answer: 1008

Solution

Let the sides be

n - d, n, n + d

(in AP) opposite to angles

α, \frac{π}{2} - 2 α, \frac{π}{2} + α

respectively (since largest - smallest =

\frac{π}{2}

).

Using sine rule with circumradius

R = 1

n - d = 2 sin α, n = 2 sin (\frac{π}{2} - 2 α) = 2 cos 2 α, n + d = 2 cos α

From the first and third:

2 n = 2 sin α + 2 cos α

, so

n = sin α + cos α

But also

n = 2 cos 2 α

, so:

2 cos 2 α = sin α + cos α

2 (cos^{2} α - sin^{2} α) = sin α + cos α

2 (cos α - sin α) = 1

Squaring:

4 (cos^{2} α + sin^{2} α - 2 sin α cos α) = 1

4 (1 - sin 2 α) = 1

sin 2 α = \frac{3}{4}

Then

cos 2 α = \pm \frac{7}{4}

. Since the triangle is obtuse,

cos 2 α = \frac{7}{4}

.

Area:

a = \frac{1}{2} \cdot n \cdot (n + d) \cdot sin α = \frac{1}{2} \cdot 2 cos 2 α \cdot 2 cos α \cdot sin α = sin 2 α \cdot cos 2 α = \frac{3}{4} \cdot \frac{7}{4} = \frac{3 7}{16}

(64 a)^{2} = (64 \cdot \frac{3 7}{16})^{2} = (127)^{2} = 144 \cdot 7 = 1008

JEE Advanced · 2023 · 04 Jun 2023 · Paper 2 · Numerical answer

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JEE Advanced · 2023 · 04 Jun 2023 · Paper 2 · Properties of Triangle

Solution

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