Samacheer Kalvi 10th Maths Solutions Chapter 6 Exercise 6.3 Pdf

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TN Samacheer Kalvi 10th Maths Solutions Chapter 6 – Trigonometry

Board	TNSCERT Class 10th Maths
Class	10
Subject	Maths
Chapter	6 (Exercise 6.3)
Chapter Name	Trigonometry

TNSCERT Class 10th Maths Pdf | all Exercise Solution

Exercise – 6.3

(1) From the top of a rock 50√3m height, the angle of depression of a car on the ground is observed to be 30°. Find the distance of the car from the rock.

Solution:

Given that AB = 50√3 m

Let, BC = x m

Now, ∆ABC,

tan30° = AB/BC

1/√3 = 50√3/x

x = 50√3 × √3

x = 50 × 3 = 150m

Thus, the distance of the car from the rock is 150m.

(2) The horizontal distance between two buildings is 70 m. The angle of depression of the top of the first building when seen from the top of the second building is 45°. If the height of the second building is 120 m, find the height of the first building.

Solution:

AB = first building and EC = 2nd building.

Let, AB = x m, EC = 120m

Now, DE = EC – DC

= EC – AB [∵ AB = DC]

= (120 – x) m.

BC = 70m (Given)

AD = BC = 70m

Now, ∆ADE,

tan45° = ED/AD

1 = (120-x)/70

70 = 120 – x

x = 120 – 70

x = 50m

∴ Now, DC = 50m.

∴ AB = 1st building height = 50m.

Thus, the height of the first building is 50m.

(3) From the top of the tower 60 m high the angles of depression of the top and bottom of a vertical lamp post are observed to be 38° and 60° respectively. Find the height of the lamp post. (tan38° = 0.7813, √3 = 1.732)

Solution:

Given that, BE = 60m

AD = Lamp post height = x m (Let)

Now, ∆ABE,

tan60° = BE/AB

√3 = 60/AB

AB = 60/√3

AB = 20√3 m

Now, EC = BE – BC

= BE – AD [∵ BC = AD]

= (60 – x) m

Now, ∆DCE,

tan38° = EC/DC

tan 38° = EC/AB [∵ DC = AB]

0.7813 = (60-x)/20√3

60 – x = 20√3 × 0.7513

60 – x = 20 × 1.732 × 0.7813

60 – x = 27.06

x = 60 – 27.06

x = 32.93m

∴ AD = x = 32.93 m

Thus, the height of the lamp post is 32.9 m.

(4) An aeroplane at an altitude of 1800 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° and 30° respectively. Find the distance between the two boats. (√3 = 1.732)

Solution:-

Given that, BD = 1800 m.

Now, ∆DCB,

tan60° = BD/BC

√3 = 1800/BC

BC = 1860/√3

BC = 600√3

Now, AB = AC + BC

= AC + 600√3

Now, ∆ABD,

tan30° = BD/AB

1/√3 = 1800/AC+600√3

AC + 600√3 = 1800√3

AC = 1800√3 – 600√3

AC = 1200√3

Now, two boats distance = AC

= AB – BC

= (AC + 600√3) – 600 √3

= 1200√3 – 600√3 – 600√3

= 1200 × 1.732 (Given)

= 2078.4 m

Thus, the distance between the two boats is 2078.4 m.

(5) From the top of a lighthouse, the angle of depression of two ships on the opposite sides of it are observed to be 30° and 60°. If the height of the lighthouse is h meters and the line joining the ships passes through the foot of the lighthouse, show that the distance between the ships is 4h/√3 m.

Solution:

AD = light house height = h m

Now, ∆ABD

tan60° = AD/BD

BD = AD/tan60°

BD = h/√3

Now, ∆ADC,

tan30° = AD/DC

1/√3 = h/DC

DC = √3 h

Now, two ships distance

= BC

= BD + DC

= h/√3 + √3 h

= h + 3h/√3

= 4h/√3

Thus, the distance between the ships is 4h/√3 m. [Proved]

(6) A lift in a building of height 90 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°. Two minutes later, the angle of depression reduces to 30°. If the fountain is 30√3 feet from the entrance of the lift, find the speed of the life which is descending.

Solution:

Given that DC = 30√3 feet and AC = 90 feet

Now, ∆BCD,

tan30° = BC/DC

1/√3 = BC/30√3

BC = 30√3/√3

BC = 30 feet

Now, AB = AC – BC

= (90 – 30) feet

= 60 feet

Now, 2 minutes lift descending 60 feet.

∴ Speed of the lift = 60/2 feet/minutes

= 30 feet/minutes

Thus, the speed of the lift 30 feet/minutes.

Here is your solution of TN Samacheer Kalvi Class 10 Math Chapter 6 Trigonometry Exercise 6.3

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