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In a ∆ABC, 2ac sin ((A – B + C)/2) is equal to

(1) a2+b2−c2 (2) c2+a2−b2 (3) b2−c2−a2 (4) c2−a2−b2 Solution: We know A + B + C = 1800 So A + C = 180 - B 2ac sin ((A - B + C)/2) = 2ac... View Article

If angles A, B and C are in AP, then (a + c)/b is equal to

1) 2 sin (A - C)/2 2) 2 cos (A - C)/2 3) cos (A - C)/2 4) sin (A - C)/2 Solution: Given A, B and C are in AP. So 2B = A + C B = (A + C)/2... View Article

Let ABC be a triangle such that ∠ACB = π/6 and let a, b and 6 c denote the lengths of the sides opposite to A, B and C respectively. The value(s) of x for which a = x² + x + 1, b = x² – 1 and c = 2x + 1 is (are)

1) -(2 + √3) 2) 1 + √3 3) 2 + √3 4) 4√3 Solution: We know cos C = (a2 + b2 - c2)/2ab cos 30 = [(x2 + x + 1)2 + (x2 - 1)2 - (2x + 1)2]/2(x2 +... View Article

Let p, q, and r be the sides opposite to the angles P, Q, R respectively in a ∆PQR. If r² sin P sin Q = pq, then the triangle is

1) equilateral 2) acute-angled but not equilateral 3) obtuse-angled 4) right-angled Solution: Given r2 sin P sin Q = pq…(i) We know in ∆PQR,... View Article

If PQR be a triangle of area ∆ with a = 2, b = 7/2 and c = 5/2, where a, b and c are the lengths of the sides of the triangle opposite to the angles at P, Q and R, respectively. Then (2 sin P – sin 2P)/(2 sin P + sin 2P) is equal to

1) 3/4∆ 2) 45/4∆ 3) (3/4∆)2 4) (45/4∆)5 Solution: Given a = 2, b = 7/2 and c = 5/2 s = (a + b + c)/2 = 4 (2 sin P - sin 2P)/(2 sin P + sin... View Article

In a triangle, if r₁ + r₃ = k cos² B/2, then k is equal to

1) R 2) 2R 3) 3R 4) 4R Solution: Given r1 + r3 = k cos2 B/2 We know r1 = 4R sin (A/2) cos (B/2) cos (C/2) r3 = 4R cos (A/2) cos (B/2) sin... View Article

In ∆ABC with usual notation, observe the two statements given below

I . rr₁r₂r₃ = ∆²

II. r₁r₂ + r₂r₃ + r₃r₁= s²

Which of the following is correct?

1) Both I and II are correct 2) I and II are incorrect 3) I is incorrect, II is correct 4) I is correct, II is correct Solution: Consider... View Article

In ∆ABC, ∠A = π/2, b = 4, c = 3, then the value of R/r is

1) 5/2 2) 7/2 3) 9/2 4) 35/24 Solution: Given ∠A = π/2, b = 4, c = 3 This is a right angled triangle. So a2 = b2 +c2 = 42 + 32 = 16 + 9 =... View Article

In a triangle (1 – r₁/r₂)(1 – r₁/r₃) = 2, then the triangle is

1) equilateral 2) right angled 3) isosceles 4) none of these Solution: In a triangle r1 = ∆/(s - a) r2 = ∆/(s - b) r3 = ∆/(s - c) Given (1 -... View Article

Two sides of a triangle are given by the roots of the equations x² – 5x + 6 = 0 and the angle between the sides is π/3. Then, the perimeter of the triangle is

1) 5 + √2 2) 5 + √3 3) 5 + √5 4) 5 + √7 Solution: Given x2 - 5x + 6 = 0 (x - 2)(x - 3) = 0 So x = 2 and x = 3 Let b = 2 and c = 3 Angle... View Article

In ∆ABC, if (a + b + c) (a – b + c) = 3ac, then

1) ∠B = 60o 2) ∠B = 30o 3) ∠C = 60o 4) ∠A + ∠C = 90o Solution: Given (a + b + c) (a - b + c) = 3ac ((a + c) + b)((a + c) - b) = 3ac (a + c)2... View Article

If a₁, a₂,….. a_n is an AP with common difference d, then tan [tan^-1 d / [1 + a₁a₂] + tan^-1 d / [1 + a₂a₃] + …. + tan^-1 d / [1 + a_n-1a_n]] =

1) [n - 1] d / [a1 + an] 2) [n - 1] d / [1 + a1an] 3) nd / [1 + a1an] 4) [an - a1] / [an + a1] Solution: (2) [n - 1] d / [1 + a1an] tan-1 d... View Article

If tan^-1 x + cos^-1 y / (√1 + y²) = sin^-1 3 / √10 and both x and y are positive and integral, then x and y =

1) (1, 2) and (2, 7) 2) (1, 2) and (1, 7) 3) (1, 7) and (2, 7) 4) (1, 7) and (2, 1) Solution: (1) (1, 2) and (2, 7) tan-1 x + tan-1 (1 / y)... View Article

If x₁, x₂, x₃, x₄ are roots of the equation x⁴ – x³ sin 2β + x² cos 2β – x cos β – sin β = 0, the tan^-1 x₁ + tan^-1 x₂ + tan^-1 x₃ + tan^-1 x₄ =

1) β 2) (π / 2) - β 3) π - β 4) - β Solution: (2) (π / 2) - β ∑x1 = sin 2β, ∑x1x2 = cos 2β, ∑x1x2x3 = cos β and x1x2x3x4 = - sin β tan-1 x1... View Article

If cos^-1 (x / a) + cos^-1 (y / b) = ɑ, then (x² / a²) – (2xy / ab) cos ɑ + (y² / b²) =

1) sin2 ɑ 2) cos2 ɑ 3) tan2 ɑ 4) cot2 ɑ Solution: (1) sin2 ɑ cos-1 [(x / a) * (y / b) - √1 - (x2 / a2) √1 - (y2 / b2)] = ɑ (xy / ab) - √1 -... View Article

If a, b, c be positive real number and the value of θ = tan^-1 √a (a + b + c) / bc + tan^-1 √b (a + b + c) / ca + tan^-1 √c (a + b + c) / ab, then tan θ =

1) 0 2) 1 3) [a + b + c] / abc 4) None of these Solution: (1) 0 θ = tan-1 √a (a + b + c) / bc + tan-1 √b (a + b + c) / ca + tan-1 √c (a + b... View Article

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