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If we consider only the principal values of the inverse trigonometric functions, then the value of tan [cos^-1 (1 / 5√2) – sin^-1 (4 / √17)] is

1) √29 / 3 2) 29 / 3 3) √3 / 29 4) 3 / 29 Solution: (4) 3 / 29 tan [cos-1 (1 / 5√2) - sin-1 (4 / √17)] = tan [tan-1 7 - tan-1 4] = tan... View Article

A triangular park is enclosed on two sides by a fence and on the third side by a straight river bank. The two sides having fences are of same length x. The maximum area enclosed by the park is

1) √(x3/8) 2) x2/2 3) πx2 4) 3x2/2 Solution: Let ABC be the triangle such that AB = AC = x And let ∠ACB = sin θ Draw AD perpendicular to BC.... View Article

Points D, and E are taken on the side BC of the ∆ABC, such that BD = DE = EC. If ∠BAD = x, ∠DAE = y and ∠EAC = z, then the value of sin(x + y) sin (y + z)/sin x sin z is

1) 1 2) 2 3) 4 4) None of these Solution: Consider the figure below. From ∆ADC, sin (y + z)/DC = sin C/AD .....(1) From ∆ABD sin x/BD =... View Article

The sides of a triangle are in the ratio 1 : √3 : 2, then the angles of the triangle are in the ratio

1) 1 : 3 : 5 2) 2 : 3 : 1 3) 3 : 2 : 1 4) 1 : 2 : 3 Solution: Given a: b : c = 1 : √3 : 2 Let a = λ, b = √3λ, c = 2λ From cosine rule cos C... View Article

In a ∆ABC, if tan A/2 = 5/6, tan C/2 = 2/5, then

1) a, c, and b are in AP 2) a, b, and c are in GP 3) b, a, and c are in AP 4) a, b, and c are in AP Solution: Given tan A/2 = 5/6, tan C/2 =... View Article

If in ∆ABC, sin A/2 sin C/2 = sin B/2 and 2s is the perimeter of the triangle, then s is equal to

1) 2b 2) b 3) 3b 4) 4b Solution: We know sin A/2 = √((s - b)(s - c)/bc) sin B/2 = √((s - c)(s - a)/ac) sin C/2 = √((s - a)(s - b)/ab) Given... View Article

In a ∆ABC , (b + c) cos A + (a + c) cos B + (a + b) cos C is equal to

(1) 0 (2) abc (3) a + b + c (4) 1 Solution: (b + c) cos A + (a + c) cos B + (a + b) cos C = b cos A + c cos A + a cos B + c cos B + a cos C... View Article

In ∆ABC, if cot A, cot B, and cot C are in AP, then a², b², and c²are in

1) GP 2) AP 3) HP 4) None of these Solution: Given cot A, cot B, and cot C are in AP. So cot B - cot A = cot C - cot B cos B/sin B - cos... View Article

If in a ∆ABC, 4 sin A = 4 sin B = 3 sin C, the cos C is equal to

1) 1/3 2) 1/9 3) 1/27 4) 1/18 Solution: Let 4 sin A = 4 sin B = 3 sin C = k sin A = k/4, sin B = k/4, sin C = k/3 From sine rule, we have... View Article

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

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