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In a ∆ABC , if (√3 – 1)a = 2b, A = 3B, then ∠C is equal to

1) 60o 2) 120o 3) 30o 4) 45o Solution: Given (√3 - 1)a = 2b So a/b = 2/(√3 - 1) Also given A = 3B By sine rule a/sin A = b/sin B So a/b =... View Article

If in a ∆ABC, 2b² = a² + c², then (sin 3B/sin B) is equal to

(1) (c2 - a2)/2ac (2) (c2 - a2)/ac (3) ((c2 - a2)/ac)2 (4) ((c2 - a2)/2ac)2 Solution: We know sin 3B = 3 sin B - 4 sin3 B sin 3B/sin B = 3 -... View Article

In ∆ABC, (a – b)² cos² C/2 + (a + b)² sin² C/2 is equal to

1) a2 2) b2 3) c2 4) None of these Solution: (a - b)2 cos2 C/2 + (a + b)2 sin2 C/2 = (a2 + b2 - 2ab) cos2 C/2 + (a2 + b2 + 2ab) sin2 C/2 =... View Article

In any ∆ABC, (tan A/2 – tan B/2)/(tan A/2 + tan B/2) is equal to

(1) (a - b)/(a + b) (2) (a - b)/c (3) c/(a - b) (4) none of these Solution: (tan A/2 - tan B/2)/(tan A/2 + tan B/2) = ((sin A/2)/(cos A/2) -... View Article

In a ∆ABC, if sin² A/2, sin² B/2 and sin² C/2 are in HP. Then, a, b and c will be in

1) AP 2) GP 3) HP 4) None of these Solution: We have sin2 A/2 = (s - b)(s- c)/bc sin2 B/2 = (s - a)(s - c)/ac sin2 C/2 = (s - a)(s - b)/ab... View Article

In a ∆ABC, 2a² + 4b² + c²= 4ab + 2ac , then cos B is

1) 0 2) 1/8 3) 3/8 4) 7/8 Solution: Given 2a2 + 4b2 + c2 = 4ab + 2ac a2 + a2 + 4b2 + c2 = 4ab + 2ac (a2 - 2ac + c2) + (a2 + 4b2 - 4ab) = 0... View Article

If a = 2√2 , b = 6, A = 45^o , then

1) no triangle is possible 2) one triangle is possible 3) two triangles are possible 4) either no triangle or two triangles are possible... View Article

In a ∆ABC, if sin A sin B = ab/c², then the triangle is

1) equilateral 2) isosceles 3) right-angled 4) obtuse-angled Solution: Given sin A sin B = ab/c2 Cross multiply, we get c2 sin A sin B = ab... View Article

In ∆ABC, a = 2, b = 3, and sin A = 2/3 , then B is equal to

1) 30o 2) 60o 3) 90o 4) 120o Solution: Given a = 2, b = 3, and sin A = 2/3 Using sine rule a/sin A = b/sin B 2/(⅔) = 3/sin B 3 = 3/sin B sin... View Article

If one side of a triangle is double the other and the angles opposite to these sides differ by 60⁰, then the triangle is

1) obtuse-angled 2) acute-angled 3) isosceles 4) right-angled Solution: Let a = 2b A - B = 600…(i) We have tan ((A - B)/2) = ((a - b)/(a +... View Article

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 of BC and AD is perpendicular to AC... 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 + 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

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