Latest Posts

The sides BC, CA, and AB of a ∆ABC are of lengths a, b, and c, respectively. If D is the mid-point of BC and AD is perpendicular to AC, then the value of cos A cos C is

(1) 2(c2 – a2)/3ac (2) 3(c2 – a2)/2ac (3) 1 (4) none of these Solution: Given D is the mid-point... View Article

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... 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... 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 =... 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)... 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... 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... 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... 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... 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)... 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 –... 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... 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... 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,... 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)... 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)... 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... View Article

«
1
…
3,028
3,029
3,030
3,031
3,032
…
5,472
»