Mastering Trigonometry for Quantitative Aptitude (Chapter-12): 100 MCQS to Ace Competitive Exams in India December 31, 2023 Quantitative AptitudeTest 1Test 2Test 3Test 4Test 5 Report a question What's wrong with this question?You cannot submit an empty report. Please add some details. 0% Trigonometric Test Questions-520 MCQs20 Marks10 Minutes 1 / 20The value of cos² x + cos² y – 2cos x × cos y × cos (x + y) is sin (x + y) sin² (x + y) sin4 (x + y) sin³ (x + y) 2 / 20If 3 × tan(x – 15) = tan(x + 15), then the value of x is 90 45 60 30 3 / 20The value of cos 20 + 2sin² 55 – √2 sin65 is -1 1 0 None of these 4 / 20The angle of elevation of a cloud from a point h metres above a lake is β. The angle of depression of its reflection in the lake is 45° . The height of location of the cloud from the lake is h tan(45° − β) h(1-tanβ)/(1-tanβ) h(1+tanβ)/(1-tanβ) h(1-tanβ)/(1+tanβ) 5 / 20If a×cos x + b × cos x = c, then the value of (a × sin x – b²cos x)² is a² + b² – c² a² – b² + c² a² – b² – c² a² + b² + c² EXPLANATION:We have (a×cos x + b × sin x)² + (a × sin x – b × cos x)² = a² + b² ⇒ c² + (a × sin x – b × cos x)² = a² + b² ⇒ (a × sin x – b × cos x)² = a² + b² – c² 6 / 20(1 + tan θ + sec θ)(1 + cot θ − cosecθ) is equal to 2/5 1/3 2 3/2 7 / 20A plane is observed to be approaching the airport. It is at a distance of 12 km from the point of observation and makes an angle of elevation of 60°. The height above the ground of the plane is 2√3 m 6√3 m 3√3 m 4√3 m 8 / 20The tops of two poles of height 20 m and 14 m are connected by a wire. If the wire makes an angle of 30° with horizontal, the length of the wire is 20 m 10 m 12 m 6 m 9 / 20If sin θ + cos θ = a and sec θ + cosec θ = b , then the value of b(a²-1) is equal to 2a 0 2ab 3a 10 / 20The value of (sin 7x + sin 5x) /(cos 7x + cos 5x) + (sin 9x + sin 3x) / (cos 9x + cos 3x) is tan 6x 4 tan 6x 2 tan 6x 3 tan 6x EXPLANATION:Given, (sin 7x + sin 5x) /(cos 7x + cos 5x) + (sin 9x + sin 3x) / (cos 9x + cos 3x) ⇒ [{2 × sin(7x+5x)/2 × cos(7x-5x)/2}/{2 × cos(7x+5x)/2 × cos(7x-5x)/2}] + [{2 × sin(9x+3x)/2 × cos(9x-3x)/2}/{2 × cos(9x+3x)/2 × cos(9x-3x)/2}] ⇒ [{2 × sin 6x × cosx}/{2 × cos 6x × cosx}] + [{2 × sin 6x × cosx}/{2 × cos 6x × cosx}] ⇒ (sin 6x/cos 6x) + (sin 6x/cos 6x) ⇒ tan 6x + tan 6x ⇒ 2 tan 6x 11 / 20If sin θ = cos θ , then 2 tan²θ + sin²θ −1 is equal to 3/2 2 1/3 2/5 12 / 20The value of cos 5π is None of these 0 -1 1 13 / 20Two persons are standing ‘x’ metres apart from each other and the height of the first person is double that of the other. If from the middle point of the line joining their feet an observer finds the angular elevations of their tops to be complementary, then the height of the shorter person (in metres) is x/(3√2) x/√2 2x x/(2√2) 14 / 20If the radius of the circumcircle of an isosceles triangle PQR is equal to PQ ( = PR), then the angle P is π/6 π/2 π/3 2π/3 EXPLANATION:Let S be the center of the circumcircle of triangle PQR. So, SP = SQ = SR = PQ = PR, where SP, SQ & SR are radii. Thus SPQ & SPR are equilateral triangles. ⇒ ∠QSP = 60°; Similarly Angle RQP = 60° ⇒ Angle at the center QSP = 120° So, SRPQ is a rhombus, since all the four sides are equal. Hence, its opposite angles are equal; so Angle P = Angle QSP = 120° 15 / 20If the ratio of the height of a tower and the length of its shadow is √3 : 1 then the angle of elevation of the sun has measure 60° 45° 33° 20° 16 / 20A tower is 60 m height. Its shadow is x metres shorter when the sun’s altitude is 45° than when it has been 30° , then x is equal to 2/5 3/2 2 1/3 17 / 20The electric pole subtends an angle of 30° at a point on the same level as its foot. At a second point ‘b’ metres above the first, the depression of the foot of the tower is 60° . The height of the tower (in metres) is equal to 23.43 33.53 49.95 43.92 18 / 20In a triangle ABC, cosec A (sin B cos C + cos B sin C) equals c/a 1 a/c none of these 19 / 20If tan A – tan B = x and cot B – cot A = y, then the value of cot (A – B) is 1/x + 1/y x + 1/y 1/x + y x + y EXPLANATION:Given, tan A – tan B = x ……………. 1 and cot B – cot A = y ……………. 2 From equation, 1/cot A – 1/cot B = x ⇒ (cot B – cot A)/(cot A × cot B) = x ⇒ y/(cot A × cot B) = x {from equation 2} ⇒ y = x × (cot A × cot B) ⇒ cot A × cot B = y/x Now, cot (A – B) = (cot A × cot B + 1)/(cot B – cot A) ⇒ cot (A – B) = (y/x + 1)/y ⇒ cot (A – B) = (y/x) × (1/y) + 1/y ⇒ cot (A – B) = 1/x + 1/y 20 / 20The value of 4 × sin x × sin(x + π/3) × sin(x + 2π/3) is sin 2x sin 4x sin x sin 3x Previous 1 2 3 4 5 6NextRelated Posts Profit and Loss MCQs (Chapter-4): 100 Essential Questions for Competitive Exams in India Permutations & Combinations Practice Test (Chapter-8): 100 MCQs for Competitive Exams in India Simplification & Approximation Questions (Chapter-9): Important MCQs to Ace Competitive ExamsLeave a Reply Cancel Reply Save my name, email, and website in this browser for the next time I comment. I have read and accepted the Privacy Policy * Δ