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Samacheer Kalvi 10th Maths Exercise 6.3 Solutions

10th Maths Exercise 6.3 solutions (Chapter 6 Trigonometry) with step-by-step answers for all 6 questions, as per the 2026-27 Samacheer Kalvi textbook.

Answers at a glance5 answers · full steps below
  1. Q1Distance of the car from the rock =150=150 m
  2. Q2Height of the first building =50=50 m
  3. Q3Height of the lamp post =32.94=32.94 m (approximately)
  4. Q4Distance between the two boats =12003=2078.4=1200\sqrt{3}=2078.4 m
  5. Q6Speed of the lift =30=30 feet/minute =0.5=0.5 feet/second

Q1. From the top of a rock 50350\sqrt{3} m high, the angle of …

From the top of a rock 50350\sqrt{3} m high, the angle of depression of a car on the ground is observed to be 30°30°. Find the distance of the car from the rock.

  1. Let ABAB be the rock of height 50350\sqrt{3} m and CC the position of the car.

  2. Angle of depression =30°=30°, so ∠ACB=30°\angle ACB = 30° (alternate angles).

  3. In right triangle ABCABC, tan⁡30°=ABBC\tan 30° = \dfrac{AB}{BC}.

  4. 13=503BC\dfrac{1}{\sqrt{3}} = \dfrac{50\sqrt{3}}{BC}

  5. BC=503×3=150BC = 50\sqrt{3}\times\sqrt{3} = 150

Answer

Distance of the car from the rock =150=150 m

Q2. The horizontal distance between two buildings is 7070 m. The angle of …

The horizontal distance between two buildings is 7070 m. The angle of depression of the top of the first building when seen from the top of the second building is 45°45°. If the height of the second building is 120120 m, find the height of the first building.

Show steps (6)

  1. Let the second building be AB=120AB = 120 m and the first building be CDCD.

  2. Horizontal distance BD=70BD = 70 m, angle of depression from AA to CC is 45°45°.

  3. Draw CE⊥ABCE \perp AB. Then CE=70CE = 70 m and ∠ACE=45°\angle ACE = 45°.

  4. In right triangle ACEACE, tan⁡45°=AECE\tan 45° = \dfrac{AE}{CE}.

  5. 1=AE70⇒AE=701 = \dfrac{AE}{70} \Rightarrow AE = 70 m

  6. Height of first building =AB−AE=120−70=50= AB - AE = 120 - 70 = 50

Answer

Height of the first building =50=50 m

Q3. From the top of the tower 6060 m high the angles of …

From the top of the tower 6060 m high the angles of depression of the top and bottom of a vertical lamp post are observed to be 38°38° and 60°60° respectively. Find the height of the lamp post. (tan⁡38°=0.7813, 3=1.732)(\tan 38° = 0.7813,\ \sqrt{3} = 1.732)

Show steps (7)

  1. Let ABAB = tower of height 6060 m and CDCD = lamp post of height hh m, at distance BD=xBD = x m.

  2. Depression of the bottom DD is 60°60°: in triangle ABDABD, tan⁡60°=60x\tan 60° = \dfrac{60}{x}.

  3. x=603=203=20(1.732)=34.64x = \dfrac{60}{\sqrt{3}} = 20\sqrt{3} = 20(1.732) = 34.64 m

  4. Draw CE⊥ABCE \perp AB, so CE=x=34.64CE = x = 34.64 m and AE=60−hAE = 60 - h.

  5. Depression of the top CC is 38°38°: tan⁡38°=60−h34.64\tan 38° = \dfrac{60-h}{34.64}.

  6. 60−h=34.64×0.7813=27.0660 - h = 34.64 \times 0.7813 = 27.06

  7. h=60−27.06=32.94h = 60 - 27.06 = 32.94

Answer

Height of the lamp post =32.94=32.94 m (approximately)

Q4. An aeroplane at an altitude of 18001800 m finds that two boats …

An aeroplane at an altitude of 18001800 m finds that two boats are sailing towards it in the same direction. The angles of depression of the boats as observed from the aeroplane are 60°60° and 30°30° respectively. Find the distance between the two boats. (3=1.732)(\sqrt{3} = 1.732)

Show steps (7)

  1. Let AA be the aeroplane at height AB=1800AB = 1800 m; boats at CC and DD on the same side.

  2. Depression of the nearer boat CC is 60°60°: tan⁡60°=1800BC\tan 60° = \dfrac{1800}{BC}.

  3. BC=18003=6003BC = \dfrac{1800}{\sqrt{3}} = 600\sqrt{3} m

  4. Depression of the farther boat DD is 30°30°: tan⁡30°=1800BD\tan 30° = \dfrac{1800}{BD}.

  5. BD=18003BD = 1800\sqrt{3} m

  6. Distance between boats =BD−BC=18003−6003=12003= BD - BC = 1800\sqrt{3} - 600\sqrt{3} = 1200\sqrt{3}

  7. =1200×1.732=2078.4= 1200 \times 1.732 = 2078.4

Answer

Distance between the two boats =12003=2078.4=1200\sqrt{3}=2078.4 m

Q5. From the top of a lighthouse, the angle of depression of two …

From the top of a lighthouse, the angle of depression of two ships on the opposite sides of it are observed to be 30°30° and 60°60°. If the height of the lighthouse is hh meters and the line joining the ships passes through the foot of the lighthouse, show that the distance between the ships is 4h3\dfrac{4h}{\sqrt{3}} m.

Show steps (8)

  1. Let ABAB be the lighthouse of height hh m with foot BB, and let the ships be at CC and DD on opposite sides of BB.

  2. Angles of depression are 30°30° and 60°60°, so ∠ACB=30°\angle ACB = 30° and ∠ADB=60°\angle ADB = 60°.

  3. In right triangle ABCABC, tan⁡30°=ABBC\tan 30° = \dfrac{AB}{BC}.

  4. 13=hBC⇒BC=h3\dfrac{1}{\sqrt{3}} = \dfrac{h}{BC} \Rightarrow BC = h\sqrt{3}

  5. In right triangle ABDABD, tan⁡60°=ABBD\tan 60° = \dfrac{AB}{BD}.

  6. 3=hBD⇒BD=h3\sqrt{3} = \dfrac{h}{BD} \Rightarrow BD = \dfrac{h}{\sqrt{3}}

  7. Since CC, BB, DD lie on a line, distance between the ships =BC+BD= BC + BD.

  8. =h3+h3=3h+h3=4h3= h\sqrt{3} + \dfrac{h}{\sqrt{3}} = \dfrac{3h + h}{\sqrt{3}} = \dfrac{4h}{\sqrt{3}}

Answer

Hence the distance between the ships is 4h3\dfrac{4h}{\sqrt{3}} m

Q6. A lift in a building of height 9090 feet with transparent glass …

A lift in a building of height 9090 feet with transparent glass walls is descending from the top of the building. At the top of the building, the angle of depression to a fountain in the garden is 60°60°. Two minutes later, the angle of depression reduces to 30°30°. If the fountain is 30330\sqrt{3} feet from the entrance of the lift, find the speed of the lift which is descending.

Show steps (7)

  1. Let ABAB be the building of height 9090 ft and FF the fountain with BF=303BF = 30\sqrt{3} ft.

  2. At the top AA, depression =60°=60°: check tan⁡60°=90303=33=3\tan 60° = \dfrac{90}{30\sqrt{3}} = \dfrac{3}{\sqrt{3}} = \sqrt{3}.

  3. Let CC be the position of the lift after 22 minutes, at height CB=xCB = x ft.

  4. Depression from CC is 30°30°: tan⁡30°=x303\tan 30° = \dfrac{x}{30\sqrt{3}}.

  5. 13=x303⇒x=30\dfrac{1}{\sqrt{3}} = \dfrac{x}{30\sqrt{3}} \Rightarrow x = 30 ft

  6. Distance descended =90−30=60= 90 - 30 = 60 ft in 22 minutes.

  7. Speed =602=30= \dfrac{60}{2} = 30 ft/minute =3060=0.5= \dfrac{30}{60} = 0.5 ft/second

Answer

Speed of the lift =30=30 feet/minute =0.5=0.5 feet/second

Quick check

  1. A tower is 60 m heigh. Its shadow reduces by xx metres when the angle of elevation of the sun increases from 30∘30^\circ to 45∘45^\circ then xx is equal to

    1. (1) 41.92 m
    2. (2) 43.92 m
    3. (3) 43 m
    4. (4) 45.6 m
    Show answer

    Answer: (2) 43.92 m

    1. Shadow at 30∘30^\circ: 60tan⁡30∘=603\dfrac{60}{\tan 30^\circ}=60\sqrt3 m.

    2. Shadow at 45∘45^\circ: 60tan⁡45∘=60\dfrac{60}{\tan 45^\circ}=60 m.

    3. x=603−60=60(1.732−1)=43.92x=60\sqrt3-60=60(1.732-1)=43.92 m.

    Question 1 of 5

  2. If sin⁡θ+cos⁡θ=a\sin\theta + \cos\theta = a and sec⁡θ+cosec θ=b\sec\theta + \mathrm{cosec}\,\theta = b, then the value of b(a2−1)b(a^2-1) is equal to

    1. (1) 2a2a
    2. (2) 3a3a
    3. (3) 00
    4. (4) 2ab2ab
    Show answer

    Answer: (1) 2a2a

    1. a2−1=(sin⁡θ+cos⁡θ)2−1=2sin⁡θcos⁡θa^2-1=(\sin\theta+\cos\theta)^2-1=2\sin\theta\cos\theta.

    2. b=sec⁡θ+cosec θ=sin⁡θ+cos⁡θsin⁡θcos⁡θ=asin⁡θcos⁡θb=\sec\theta+\mathrm{cosec}\,\theta=\dfrac{\sin\theta+\cos\theta}{\sin\theta\cos\theta}=\dfrac{a}{\sin\theta\cos\theta}.

    3. b(a2−1)=asin⁡θcos⁡θ×2sin⁡θcos⁡θ=2ab(a^2-1)=\dfrac{a}{\sin\theta\cos\theta}\times 2\sin\theta\cos\theta=2a.

    Question 2 of 5

  3. acot⁡θ+b cosec θ=pa\cot\theta + b\,\mathrm{cosec}\,\theta = p and bcot⁡θ+a cosec θ=qb\cot\theta + a\,\mathrm{cosec}\,\theta = q then p2−q2p^2 - q^2 is equal to

    1. (1) a2−b2a^2 - b^2
    2. (2) b2−a2b^2 - a^2
    3. (3) a2+b2a^2 + b^2
    4. (4) b−ab - a
    Show answer

    Answer: (2) b2−a2b^2 - a^2

    1. p2−q2=(acot⁡θ+b cosec θ)2−(bcot⁡θ+a cosec θ)2p^2-q^2=(a\cot\theta+b\,\mathrm{cosec}\,\theta)^2-(b\cot\theta+a\,\mathrm{cosec}\,\theta)^2.

    2. =(a2−b2)cot⁡2θ−(a2−b2)cosec2θ=(a^2-b^2)\cot^2\theta-(a^2-b^2)\mathrm{cosec}^2\theta (the 2ab2ab terms cancel).

    3. =(a2−b2)(cot⁡2θ−cosec2θ)=(a2−b2)(−1)=b2−a2=(a^2-b^2)(\cot^2\theta-\mathrm{cosec}^2\theta)=(a^2-b^2)(-1)=b^2-a^2.

    Question 3 of 5

  4. If the ratio of the height of a tower and the length of its shadow is 3:1\sqrt{3}:1, then the angle of elevation of the sun has measure

    1. (1) 45∘45^\circ
    2. (2) 30∘30^\circ
    3. (3) 90∘90^\circ
    4. (4) 60∘60^\circ
    Show answer

    Answer: (4) 60∘60^\circ

    1. tan⁡θ=heightshadow=31=3\tan\theta=\dfrac{\text{height}}{\text{shadow}}=\dfrac{\sqrt3}{1}=\sqrt3.

    2. θ=60∘\theta=60^\circ.

    Question 4 of 5

  5. tan⁡θ cosec2θ−tan⁡θ\tan\theta\,\mathrm{cosec}^2\theta - \tan\theta is equal to

    1. (1) sec⁡θ\sec\theta
    2. (2) cot⁡2θ\cot^2\theta
    3. (3) sin⁡θ\sin\theta
    4. (4) cot⁡θ\cot\theta
    Show answer

    Answer: (4) cot⁡θ\cot\theta

    1. tan⁡θ cosec2θ−tan⁡θ=tan⁡θ(cosec2θ−1)\tan\theta\,\mathrm{cosec}^2\theta-\tan\theta=\tan\theta(\mathrm{cosec}^2\theta-1).

    2. cosec2θ−1=cot⁡2θ\mathrm{cosec}^2\theta-1=\cot^2\theta.

    3. =tan⁡θ⋅cot⁡2θ=cot⁡θ=\tan\theta\cdot\cot^2\theta=\cot\theta.

    Question 5 of 5
Chapter 6 formulas
  • Pythagorean identity

    sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1

  • Identity with tangent and secant

    1+tan⁡2θ=sec⁡2θ1 + \tan^2\theta = \sec^2\theta

  • Identity with cotangent and cosecant

    1+cot⁡2θ=cosec⁡2θ1 + \cot^2\theta = \operatorname{cosec}^2\theta

  • Trigonometric identity (definition)

    An equation in trigonometric ratios that is true for all values of the angle\text{An equation in trigonometric ratios that is true for all values of the angle}

  • Line of sight

    The line drawn from the eye of an observer to the point on the object being viewed\text{The line drawn from the eye of an observer to the point on the object being viewed}

  • Angle of elevation

    Angle between the line of sight and the horizontal, when the object is above the horizontal level\text{Angle between the line of sight and the horizontal, when the object is above the horizontal level}

  • Angle of depression

    Angle between the line of sight and the horizontal, when the object is below the horizontal level\text{Angle between the line of sight and the horizontal, when the object is below the horizontal level}

  • Use of trigonometric ratios

    Height or length of an object, or distance between two far objects, can be found using trigonometric ratios\text{Height or length of an object, or distance between two far objects, can be found using trigonometric ratios}

Answers checked against the Samacheer Kalvi textbook · Updated

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