Archives

The sides of a triangle are respectively 7 cm, 4√3 cm and √13 cm, then the smallest angle of triangle is

1) π/6 2) π/3 3) π/4 4) π/5 Solution: Given sides of a triangle are respectively 7 cm, 4√3 cm and √13 cm. Here smallest side = √13 We know... View Article

In any ∆ABC, the simplified form of (cos 2A/a²) – (cos 2B/b²) is

1) a2 - b2 2) 1/(a2 - b2) 3) 1/a2 - 1/b2 4) a2 + b2 Solution: We know cos 2θ = 1 - 2 sin2 θ (cos 2A/a2) - (cos 2B/b2) = (1 - 2 sin2 A)/a2 -... View Article

If the sides of a right angled triangle form an AP, then the ‘sin’ of the acute angles are

(1) ⅗, ⅘ (2) ⅕, ⅖ (3) √3, 1/√3 (4) none of these Solution: Let the sides of the triangle be (a - d), a, (a + d) Applying pythagoras theorem... View Article

If the angles A, B and C of a triangle are in an arithmetic progression and if a, b and c denote the lengths of the sides opposite to A, B and C respectively, then the value of the expression (a/c) sin 2C + (c/a) sin 2A is

1) 1/2 2) √3/2 3) 1 4) √3 Solution: Given angles of a triangle ABC are in AP. Let a-d, a, a+d represent angles A, B and C. a-d + a + a+d =... View Article

In a ∆ABC, a = 8 cm, b = 10 cm, c = 12 cm. Then, relation between angles of the triangle is

1) C = A + B 2) C = 2B 3) C = 2A 4) C = 3A Solution: Given a = 8 cm, b = 10 cm, c = 12 cm We know cos C = (a2 + b2 - c2)/2ab = (82 + 102 -... View Article

If in a ∆ABC, sin A, sin B, and sin C are in AP, then

1) the radius are in AP 2) the altitudes are in HP 3) the angles are in AP 4) the angles are in HP Solution: Given sin A, sin B, and sin C... View Article

If in a ∆ABC , a/cos A = b/cos B, then

1) sin2 A + sin2 B = sin2 C 2) 2 sin A cos B = sin C 3) 2 sin A sin B sin C = 1 4) None of the above Solution: Given a/cos A = b/cos B a cos... View Article

If in a ∆ABC, a tan A + b tan B = (a + b) tan (A + B)/2 , then

1) A = B = C 2) C = A 3) A = B 4) B = C Solution: Given a tan A + b tan B = (a + b) tan (A + B)/2 a(tan A - tan (A + B)/2) = b(tan (A + B)/2... View Article

Angles A, B, C of a ∆ABC are in AP and b : c = √3 : √2, then ∠A is given by

1) 45o 2) 60o 3) 75o 4) 90o Solution: Given angles of a triangle ABC are in AP. Let a-d, a, a+d represent angles A, B and C. a-d + a + a+d =... View Article

If the angles of a triangle are in the ratio 1 : 1 : 4, then the ratio of the perimeter of the triangle to its largest side is

1) √2 + 2 : √3 2) 3 : 2 3) √3 + 2 : √2 4) 2 +√3 : √3 Solution: Given the angles of a triangle are in the ratio 1 : 1 : 4. Let A, B, C be the... View Article

In any ∆ABC, (a + b + c)(b + c – a)(c + a – b)(a + b – c)/4b²c² equals

1) sin2 B 2) cos2 A 3) cos2 B 4) sin2 A Solution: In ∆ABC, (a + b + c) = 2s (a + b + c)(b + c - a)(c + a - b)(a + b - c)/4b2c2 = 2s(2s - 2a)... View Article

In a ∆ABC, a, b, c are the sides of the triangle opposite to the angles A, B, C, respectively. Then, the value of a³sin (B – C) + b³ (C – A) + c³ (A – B) is

1) 0 2) 1 3) 3 4) 2 Solution: By sine rule a/sin A = b/sin B = c/sin C = k a = k sin A, b = k sin B, c = k sin C a3 = k3 sin3 A = k3 sin3 [π... View Article

The greatest and the least values of (sin^-1 x)³ + (cos^-1 x)³ are

1) (- π / 2, π / 2) 2) (- π3 / 8, π3 / 8) 3) (7π3 / 8, π3 / 32) 4) None of these Solution: (3) (7π3 / 8, π3 / 32) f (x) = (sin-1 x)3 +... View Article

The triangle formed by the tangent to the curve f(x) = x² + bx – b at the point (1, 1) and the co-ordinate axes, lies in the first quadrant. If its area is 2 then the value of b is

1) –1 2) 3 3) –3 4) 1 Solution: Given f(x) = x2 + bx - b Slope, m = dy/dx = 2x + b At (1,1) m = 2 + b The equation of the tangent at (1,1)... View Article

If the extremities of the base of an isosceles triangle are the points (2a,0) and (0,a) and the equation of one of the sides is x = 2a, then the area of the triangle is

1) 5a2 2) 5a2/2 3) 25a2/2 4) None of these Solution: Let A = (2a, 0) and B = (0, a). Let the third point on line x = 2a be C(2a, t). BC... View Article

M	T	W	T	F	S	S
					1	2
3	4	5	6	7	8	9
10	11	12	13	14	15	16
17	18	19	20	21	22	23
24	25	26	27	28	29	30
31