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NCERT Exemplar Solutions For Class 10 Maths Chapter 8Introduction To Trigonometry And Its Equations, , Exercise 8.1, Choose the correct answer from the given four options:, 1. If cos A = 4/5 , then the value of tan A is, (A) 3/5, (B) ¾, (C) 4/3, Solution:, According to the question,, cos A = 4/5 …(1), We know,, tan A = sinA/cosA, To find the value of sin A,, We have the equation,, sin2 θ +cos2 θ =1, So, sin θ = √ (1-cos2 θ), Then,, sin A = √ (1-cos2 A) …(2), sin2 A = 1-cos2 A, sin A = √(1-cos2 A), Substituting equation (1) in (2),, We get,, Sin A = √(1-(4/5)2), = √(1-(16/25)), = √(9/25), =¾, Therefore,, , 2. If sin A = ½ , then the value of cot A is, (A) √3 (B) 1/√3 (C) √3/2 (D) 1, Solution:, According to the question,, Sin A = ½ … (1), We know that,, … (2), To find the value of cos A., We have the equation,, sin2 θ +cos2 θ =1, So, cos θ = √(1-sin2 θ), Then,, cos A = √(1-sin2 A) … (3), cos2 A = 1-sin2 A, cos A = √ (1-sin2 A), Substituting equation 1 in 3, we get,, , Page No: 89, (D) 5/3

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NCERT Exemplar Solutions For Class 10 Maths Chapter 8Introduction To Trigonometry And Its Equations, 6. The value of (tan1° tan2° tan3° ... tan89°) is, (A) 0, (B) 1, (C) 2, (D) ½, Solution:, tan 1°. tan 2°.tan 3° …… tan 89°, = tan1°.tan 2°.tan 3°…tan 43°.tan 44°.tan 45°.tan 46°.tan 47°…tan 87°.tan 88°.tan 89°, Since, tan 45° = 1,, = tan1°.tan 2°.tan 3°…tan 43°.tan 44°.1.tan 46°.tan 47°…tan 87°.tan 88°.tan 89°, = tan1°.tan 2°.tan 3°…tan 43°.tan 44°.1.tan(90°-44°).tan(90°-43°)…tan(90°-3°). tan(90°2°).tan(90°-1°), Since, tan(90°-θ) = cot θ,, = tan1°.tan 2°.tan 3°…tan 43°.tan 44°.1.cot 44°.cot 43°…cot 3°.cot 2°.cot 1°, Since, tan θ = (1/cot θ), = tan1°.tan 2°.tan 3°…tan 43°.tan 44°.1. (1/tan 44o). (1/tan 43o)… (1/tan 3o). (1/tan 2o). (1/tan 1o), =1, Hence, tan 1°.tan 2°.tan 3° …… tan 89° = 1, 7. If cos 9α = sinα and 9α < 90° , then the value of tan5α is, (A) 1/√3, (B) √3, (C) 1, (D) 0, Solution:, According to the question,, cos 9∝ = sin ∝ and 9∝<90°, i.e. 9α is an acute angle, We know that,, sin(90°-θ) = cos θ, So,, cos 9∝ = sin (90°-∝), Since, cos 9∝ = sin(90°-9∝) and sin(90°-∝) = sin∝, Thus, sin (90°-9∝) = sin∝, 90°-9∝ =∝, 10∝ = 90°, ∝ = 9°, Substituting ∝ = 9° in tan 5∝, we get,, tan 5∝ = tan (5×9) = tan 45° = 1, ∴, tan 5∝ = 1

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NCERT Exemplar Solutions For Class 10 Maths Chapter 8Introduction To Trigonometry And Its Equations, , Exercise 8.2, Write ‘True’ or ‘False’ and justify your answer in each of the following:, 1. tan 47o/cot 43 ° = 1, Solution:, True, Justification:, Since, tan (90° -θ) = cot θ, , 2. The value of the expression (cos223° – sin267°) is positive., Solution:, False, Justification:, Since, (a2-b2) = (a+b)(a-b), cos2 23° - sin2 67° =(cos 23°+sin 67°)(cos 23°-sin 67°), = [cos 23°+sin(90°-23°)] [cos 23°-sin(90°-23°)], = (cos 23°+cos 23°)(cos 23°-cos 23°) (∵sin(90°-θ) = cos θ), = (cos 23°+cos 23°).0, = 0, which is neither positive nor negative, 3. The value of the expression (sin 80° – cos 80°) is negative., Solution:, False, Justification:, We know that,, sin θ increases when 0° ≤ θ ≤ 90°, cos θ decreases when 0° ≤ θ ≤ 90°, And (sin 80°-cos 80°) = (increasing value-decreasing value), = a positive value., Therefore, (sin 80°-cos 80°) > 0., 4. √((1– cos2θ) sec2 θ)= tan θ, Solution:, True, Justification:, , Page No: 93

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NCERT Exemplar Solutions For Class 10 Maths Chapter 8Introduction To Trigonometry And Its Equations, , = R.H.S, Hence proved., 3. If tan A = ¾, then sinA cosA = 12/25, Solution:, According to the question,, tan A = ¾, We know,, tan A = perpendicular/ base, So,, tan A = 3k/4k, Where,, Perpendicular = 3k, Base = 4k, , Using Pythagoras Theorem,, (hypotenuse)2 = (perpendicular)2 + (base)2, (hypotenuse)2 = (3k)2+ (4k)2 = 9k2+16k2 = 25k2, hypotenuse = 5k, To find sin A and cos A,

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NCERT Exemplar Solutions For Class 10 Maths Chapter 8Introduction To Trigonometry And Its Equations, , Hence, proved., 4. (sin α + cos α) (tan α + cot α) = sec α + cosec α, Solution:, L.H.S:, (sin α + cos α) (tan α + cot α), As we know,, , = R.H.S, Hence, proved., 5. (√3+1) (3 – cot 30°) = tan3 60° – 2 sin 60°, Solution:, L.H.S: (√3 + 1) (3 – cot 30°), = (√3 + 1) (3 – √3) [∵cos 30° = √3], = (√3 + 1) √3 (√3 - 1) [∵(3 – √3) = √3 (√3 - 1)], = ((√3)2– 1) √3 [∵ (√3+1)(√3-1) = ((√3)2 – 1)], = (3-1) √3, = 2√3, Similarly solving R.H.S: tan3 60° - 2 sin 60°, Since, tan 60o = √3 and sin 60o = √3/2,, We get,

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NCERT Exemplar Solutions For Class 10 Maths Chapter 8Introduction To Trigonometry And Its Equations, , = R.H.S, Hence, proved., 3. The angle of elevation of the top of a tower from certain point is 30°. If the observer moves 20, metres towards the tower, the angle of elevation of the top increases by 15°. Find the height of the, tower., Solution:, Let PR = h meter, be the height of the tower., The observer is standing at point Q such that, the distance between the observer and tower is QR, = (20+x) m, where, QR = QS + SR = 20 + x, ∠PQR = 30°, ∠ PSR = θ

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NCERT Exemplar Solutions For Class 10 Maths Chapter 8Introduction To Trigonometry And Its Equations, , In ∆PQR,, , Rearranging the terms,, We get 20 +x = √3h, ⇒ x = √3h – 20 …eq.1, In ∆PSR,, tan θ = h/x, Since, angle of elevation increases by 15o when the observer moves 20 m towards the tower., We have,, θ = 30° + 15° = 45°, So,, tan 45o = h/x, ⇒ 1 = h/x, ⇒h=x, Substituting x=h in eq. 1, we get, h = √3 h – 20, ⇒ √3 h – h = 20, ⇒ h (√3 - 1) = 20, , = 10 (√3 + 1), Hence, the required height of the tower is 10 (√3 + 1) meter.

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NCERT Exemplar Solutions For Class 10 Maths Chapter 8Introduction To Trigonometry And Its Equations, Solution:, Let BC = s; PC = t, Let height of the tower be AB = h., ∠ABC = θ and ∠APC = 90° - θ, (∵ the angle of elevation of the top of the tower from two points P and B are complementary), , ⇒ h2 = st, ⇒ h = √st, Hence the height of the tower is √st., 7. The shadow of a tower standing on a level plane is found to be 50 m longer when Sun’s elevation, is 30° than when it is 60°. Find the height of the tower., Solution:, Let SQ = h be the tower., ∠SPQ = 30° and ∠SRQ = 60°, According to the question, the length of shadow is 50 m long hen angle of elevation of the sun is, 30° than when it was 60°. So,, PR = 50 m and RQ = x m

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NCERT Exemplar Solutions For Class 10 Maths Chapter 8Introduction To Trigonometry And Its Equations, , So in ∆SRQ, we have, , ⇒ 50√3+h = 3h, ⇒ 50√3 = 3h – h, ⇒ 3h - h = 50√3, ⇒ 2h = 50√3, ⇒ h = (50√3)/2, ⇒ h = 25√3, Hence, the required height is 25√3 m., 8. A vertical tower stands on a horizontal plane and is surmounted by a vertical flag staff of height

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NCERT Exemplar Solutions For Class 10 Maths Chapter 8Introduction To Trigonometry And Its Equations, h. At a point on the plane, the angles of elevation of the bottom and the top of the flag staff are α, and β, respectively. Prove that the height of the tower is [h tan α/(tan β – tan α)]., Solution:, Given that a vertical flag staff of height h is surmounted on a vertical tower of height H(say),, such that FP = h and FO = H., The angle of elevation of the bottom and top of the flag staff on the plane is ∠PRO =, α and ∠FRO = β respectively, , In ∆PRO, we have, , Hence, proved., 9. If tanθ + secθ = l, then prove that secθ = (l2 + 1)/2l., Solution:, Given: tan θ+ sec θ = l …eq. 1