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If R and r are the radii of the circumcircle and incircle of a regular polygon of n sides, each side being of length a, then a is equal to

(1) 2(R + r) sin (π/2n) (2) 2(R + r) tan (π/2n) (3) 2(R + r) (4) none of these Solution: We know R = ½ a cosec (π/n) r = ½ a cot (π/n) So... View Article

If α, β are the acute angles of an obtuse-angled triangle, then tan α tan β is

1) < 1 2) > 1 3) = 1 4) None of these Solution: Given α, β are the acute angles of an obtuse-angled triangle. So α + β is also an... View Article

In a triangle ABC, if cos A + cos B + cos C = 7/4, then R/r is equal to

(1) 4/7 (2) 4/3 (3) 3/4 (4) none of these Solution: Given cos A + cos B + cos C = 7/4 If A, B, C are the angles of a triangle, then cos A +... View Article

If d₁, d₂, d₃ are the diameters of the three escribed circles of a triangle, then d₁d₂ + d₂d₃ + d₃d₁ is equal to

(1) 4s2 (2) ∆2 (3) 4∆2 (4) 2∆2 Solution: We know d1 = 2r1, d2 = 2r2,d3 = 2r3 We have r1 = ∆/(s - a) r2 = ∆/(s - b) r3 = ∆/(s - c) d1d2 +... View Article

If in a triangle ABC, c⁴ – 2c²(a² + b²) + a⁴ + a²b² + b⁴ = 0, then the acute angle C is equal to

(1) 300 (2) 600 (3) 450 (4) 750 Solution: From cosine rule, cos C = (a2 + b2 - c2)/2ab cos2 C = (a2 + b2 - c2)2/4a2b2 ..(i) Given c4 -... View Article

The sides of a triangle are in the ratio 5: 8: 11. If θ denotes the angle opposite to the largest side of the triangle, then the value of tan²θ/2 is equal to

(1) 1/21 (2) 7/48 (3) 7/3 (4) 7/12 Solution: Given the sides are in the ratio 5: 8: 11 Let the sides be 5x, 8x, and 11x. Then s = (5x + 8x +... View Article

If a⁴ + b⁴ + c⁴ = 2c²(a² + b²), then acute value of ∠C is equal to

(1) 300 (2) 600 (3) 450 (4) 750 Solution: Given a4 + b4 + c4 = 2c2 (a2 + b2) a4 + b4 + c4 - 2c2a2 - 2c2b2 = 0 Adding 2a2b2 on both sides, we... View Article

In a triangle ABC, if cos A/a = tan C/c, then sin (B + C) is equal to

(1) cos B sin A (2) cos A cos C (3) cos A sin B (4) sin B sin C Solution: Given cos A/a = tan C/c cos A/a = (sin C/cos C)×(1/c) cos A/a =... View Article

In a triangle ABC, if tan (A/2) = p, tan (B/2) = q, then 2(p + q)(1 – pq)/(1 + p²)(1 + q²) is equal to

(1) sin C (2) sin 2C (3) sin A + sin B (4) none of these Solution: Given tan (A/2) = p, tan (B/2) = q 2(p + q)(1 - pq)/(1 + p2)(1 + q2) = 2... View Article

If in a triangle ABC, the line joining the circumcentre O and the incentre I is parallel to BC, then

1) r = R cos A 2) r = R sin A 3) R = r cos A 4) R = r sin A Solution: Let I denote the incentre and O denotes the circumcentre. OI is... View Article

In a triangle, ABC, 1 – tan (A/2) tan (B/2) is equal to

(1) 2c/(a + b + c) (2) 2c/(a + b - c) (3) 2b/(a + b + c) (4) none of these Solution: We know tan(A/2) = √[(s - b)(s - c)/s(s - a)] tan (B/2)... View Article

For a regular polygon, let r and R be the radii of the inscribed and the circumscribed circles. An incorrect statement among the following is

1) there is a regular polygon with r/R = 1/2 2) there is a regular polygon with r/R = 1/√2 3) there is a regular polygon with r/R = 2/3 4)... View Article

In a ∆ABC, let ∠C = π/2, if r is the inradius and R is the circumradius of the ∆ABC, then 2 (r + R) equals

1) c + a 2) a + b + c 3) a + b 4) b + c Solution: Given ∠C = π/2 Let r is the inradius and R is the circumradius of the ∆ABC. The triangle... View Article

If p₁, p₂, p₃ are respectively the perpendiculars from the vertices of a triangle to the opposite side, then p₁p₂p₃ is equal to

1) a2b2c2 2) 2 a2b2c2 3) 4 a2b2c3/R2 4) a2b2c2/8R3 Solution: Let p1 be the altitude AD, p2 be the altitude BE and p3 be the altitude CF.... View Article

Inradius of a circle which is inscribed in an isosceles triangle one of whose angle is 2π/ 3, is √3, then the area of the triangle is

1) 4√3 2) 12 - 7√3 3) 12 + 7√3 4) None of these Solution: Given the triangle is isosceles with one angle 1200. Let A = 1200 Then b = c B =... View Article

The area of a circle and the area of a regular polygon of n sides of the perimeter equal to that of that circle are in the ratio

1) tan π/n : π/n 2) cos π/n : π/n 3) sin π/n : π/n 4) cot π/n : π/n Solution: Let r be the radius of the circle and a be the side of... View Article

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