KSEEB SSLC Class 10 Maths Solutions Chapter 12 Some Applications of Trigonometry Ex 12.1

KSEEB SSLC Class 10 Maths Solutions Chapter 12 Some Applications of Trigonometry Ex 12.1 are part of KSEEB SSLC Class 10 Maths Solutions. Here we have given Karnataka SSLC Class 10 Maths Solutions Chapter 12 Some Applications of Trigonometry Exercise 12.1.

Karnataka SSLC Class 10 Maths Solutions Chapter 12 Some Applications of Trigonometry Exercise 12.1

Question 1.
A circus artist is climbing a 20 m long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground level is 30°. (see the fig. given)
KSEEB SSLC Class 10 Maths Solutions Chapter 12 Some Applications of Trigonometry Ex 12.1 Q 1
Solution:
Length of rope is 20 m.
Height of the pole = ?, AB = ?
Let the height of the pole AB be ‘A’.
Length of rope AC = l = 20 m.
Angle of elevation, ∠BCA = 30°.
∴ Let ⊥∆AABC, ∠B = 90°.
KSEEB SSLC Class 10 Maths Solutions Chapter 12 Some Applications of Trigonometry Ex 12.1 Q 1.1
∴ AB = 10 m.
∴ Height of the pole 10 m.

KSEEB Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.4

Question 2.
A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle 30° with it. The distance between the foot of the tree to the point where the top touches the ground is 8 m. Find the height of the tree.
Solution:
Let the tree is broken at A and its top is touching the ground at B.
Now, in right ∆AOB, we have
\(\frac{A O}{O B}\) = tan 30°
KSEEB SSLC Class 10 Maths Solutions Chapter 12 Some Applications of Trigonometry Ex 12.1 Q2

Question 3.
A contractor plans to install two slides for the children to play in a park. For the children below the age of 5 years, she prefers to have a slide whose top is at a height of 1.5 m, and is inclined at an angle of 30° to the ground, whereas for elder children, she wants to have a steep slide at a height of 3 m, and inclined at an angle of 60° to the ground. What should be the length of the slide in each case ?
KSEEB SSLC Class 10 Maths Solutions Chapter 12 Some Applications of Trigonometry Ex 12.1 Q 3
Solution:
i) AB = 1.5 m., AC = l1 m., ∠ACB = 30°
In ⊥∆ABC, ∠B = 90°
KSEEB SSLC Class 10 Maths Solutions Chapter 12 Some Applications of Trigonometry Ex 12.1 Q 3.1
∴ l1 = 2 × 1.5 = 3 m
∴ Length of slide of younger children = 3 m
ii) PQ = 3 m, OP = 12 m, ∠POQ = 60°
In ⊥∆PQR, ∠Q = 90°
KSEEB SSLC Class 10 Maths Solutions Chapter 12 Some Applications of Trigonometry Ex 12.1 Q 3.2
∴ Length of slide of elder children = \(2 \sqrt{3}\) m

Question 4.
The angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of the tower, is 30°. Find the height of the tower.
Solution:
In right ∆ABC, AB = height of the tower and point C is 30 m away from the foot of the tower,
MP Board Class 10th Maths Solutions Chapter 9 Some Applications of Trigonometry Ex 9.1 6

KSEEB Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.4

Question 5.
A kite is flying at a height of 60m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 60°. Find the length of the string, assuming that there is no slack in the string.
KSEEB SSLC Class 10 Maths Solutions Chapter 12 Some Applications of Trigonometry Ex 12.1 Q 5
Solution:
P is the position of Kite. It is flying at a height of 60 m. above the ground.
The inclination of the string with the ground, ∠POQ = 60°
Let length of string be OP = L
KSEEB SSLC Class 10 Maths Solutions Chapter 12 Some Applications of Trigonometry Ex 12.1 Q 5.1
∴ Length of slide of elder children \(40 \sqrt{3}\) m.

Question 6.
A 1.5 m tall boy is standing at some distance from a 30 m tall building.The angle of elevation from his eyes to the top of the building increases from 30° to 60° as he walks towards the building. Find the distance he walked towards the building.
Solution:
Here, OA is the building.
KSEEB SSLC Class 10 Maths Solutions Chapter 12 Some Applications of Trigonometry Ex 12.1 Q4

KSEEB Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.4

Question 7.
From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a 20 m high building are 45° and 60° respectively. Find the height of the tower.
KSEEB SSLC Class 10 Maths Solutions Chapter 12 Some Applications of Trigonometry Ex 12.1 Q 7
Solution:
Let the height of building, PQ = 20 m.
PR= h m.
Bottom of transmission tower, P, vertex point ‘R’.
∠POQ = 45°, ∠ROQ = 60°.
KSEEB SSLC Class 10 Maths Solutions Chapter 12 Some Applications of Trigonometry Ex 12.1 Q 7.1
KSEEB SSLC Class 10 Maths Solutions Chapter 12 Some Applications of Trigonometry Ex 12.1 Q 7.2
∴ Height of transmission tower is = \(20(\sqrt{3}-1)\) m

Question 8.
A statue, 1.6 m tall, stands on the top of a pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and from the same point the angle of elevation of the top of the pedestal is 45°. Find the height of the pedestal.
KSEEB SSLC Class 10 Maths Solutions Chapter 12 Some Applications of Trigonometry Ex 12.1 Q 8
Solution:
Height of pedestal, PQ = h m.
PR = 1.6 m.
∠POQ = 45°
∠ROQ = 60°
KSEEB SSLC Class 10 Maths Solutions Chapter 12 Some Applications of Trigonometry Ex 12.1 Q 8.1
KSEEB SSLC Class 10 Maths Solutions Chapter 12 Some Applications of Trigonometry Ex 12.1 Q 8.2
∴ Height of pedestal, h = \(0.8 \times(\sqrt{3}+1)\) m

KSEEB Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.4

Question 9.
The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.
Solution:
In the figure, let height of the building = AB = h m
KSEEB SSLC Class 10 Maths Solutions Chapter 12 Some Applications of Trigonometry Ex 12.1 Q9
MP Board Class 10th Maths Solutions Chapter 9 Some Applications of Trigonometry Ex 9.1 12

Question 10.
Two poles of equal heights are standing opposite each other on either side of the road, which is 80 m wide. From a point between them on the road, the angles of elevation of the top of the poles are 60° and 30°, respectively. Find the height of the poles and the distances of the point from the poles.
KSEEB SSLC Class 10 Maths Solutions Chapter 12 Some Applications of Trigonometry Ex 12.1 Q 10
Solution:
Height of AB and PQ be h m.
Breadth of Road, BQ = 80 m.
‘O’ is the point on road.
∠POQ = 60°, ∠AOB = 30°
OQ = ‘x’ m. OB = ‘y’ m.
x + y = 80 m.
KSEEB SSLC Class 10 Maths Solutions Chapter 12 Some Applications of Trigonometry Ex 12.1 Q 10.1
∴ Height of pole, AB = PQ = h = \(20 \sqrt{3}\) m.
Distance of the points c from the poles 20 m. and 60 m.

KSEEB Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.4

Question 11.
A TV tower stands vertically on a bank Of a canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is 60°. From another point 20 away from this point on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is 30° (see the figure given). Find the height of the tower and the width of the canal.
KSEEB SSLC Class 10 Maths Solutions Chapter 12 Some Applications of Trigonometry Ex 12.1 Q 11
Solution:
Let the height of TV tower, AB = h m.
Breadth of a canal, BC = x m.
∠ACB = 60° ∠ADB = 30°,
DC = 20 m..
KSEEB SSLC Class 10 Maths Solutions Chapter 12 Some Applications of Trigonometry Ex 12.1 Q 11.1
From eqtn. (ii).
\(\mathrm{h}=\mathrm{x} \sqrt{3}=10 \sqrt{3}\)
∴ Height of tower = \(10 \sqrt{3}\) m
Breadth of canal = 10 m.

Question 12.
From the top of a 7 m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the height of the tower.
Solution:
In the figure, let AB be the height of the building.
∴ AB = 7 metres.
Let BC = X m = AE
MP Board Class 10th Maths Solutions Chapter 9 Some Applications of Trigonometry Ex 9.1 17
Let CD be the height of the cable tower and DE = h m
∴ In right ∆DAE, we have
MP Board Class 10th Maths Solutions Chapter 9 Some Applications of Trigonometry Ex 9.1 18

KSEEB Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.4

Question 13.
As observed from the top of a 75 m. high lighthouse from the sea level, the angles of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.
KSEEB SSLC Class 10 Maths Solutions Chapter 12 Some Applications of Trigonometry Ex 12.1 Q 13
Solution:
Height of lighthouse, PQ = 75 m.
A and B are positions of ship.
Px is Horizontal line.
∠APx = 30° and ∠BPx = 45°
∠PAQ = 30° and ∠PBQ = 45°
Let AB = x m, BQ = y m.
Then AQ = AB + BQ = (x + y) m
KSEEB SSLC Class 10 Maths Solutions Chapter 12 Some Applications of Trigonometry Ex 12.1 Q 13.1
∴ Distance between the two ships, AQ = x + y
= \(75 \sqrt{3}\) m.

KSEEB Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.4

Question 14.
A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 m from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is 60°. After some time, the angle of elevation reduces to 30° (see the figure given below). Find the distance travelled by the balloon during the interval.
KSEEB SSLC Class 10 Maths Solutions Chapter 12 Some Applications of Trigonometry Ex 12.1 Q 14
Solution:
Height of the girl, OO’ = 1.2 m
∠AOB = 60°
∠POQ = 30°
Let OB = x m, BQ = d m, O’Q’ = y m.
AB = PQ = Q’P – Q’Q
= 88.2 – O’O
= 88.2 – 1.2 = 87 m.
Let OQ = ‘y’.
Distance balloon travelled, d = BQ
= (y – x)
KSEEB SSLC Class 10 Maths Solutions Chapter 12 Some Applications of Trigonometry Ex 12.1 Q 14.1

Question 15.
A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of 30°, which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be 60°. Find the time taken by the car to reach the foot of the tower from this point.
KSEEB SSLC Class 10 Maths Solutions Chapter 12 Some Applications of Trigonometry Ex 12.1 Q 15
Solution:
Let the height of tower, PQ = h m.
Px = Horizontal line
Let the positions of 1st car and 2nd car be A and B.
∠APx = 30° ∴ ∠PAQ = 30°
∠BPx = 60° ∠PBQ = 60°
Let speed of car be x m/s.
∴ Distance, AB = 6x m.
Time required to travel from B to Q be ‘n’ seconds,
∴ Distance, BQ = nx m
KSEEB SSLC Class 10 Maths Solutions Chapter 12 Some Applications of Trigonometry Ex 12.1 Q 15.1
\(n+6=n \sqrt{3} \times \sqrt{3}\)
3n = n+6
2n=6
∴ n = 3 seconds

KSEEB Solutions for Class 10 Maths Chapter 7 Coordinate Geometry Ex 7.4

Question 16.
The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is 6 m.
Solution:
Let the height of the tower be represented by AB in the figure.
MP Board Class 10th Maths Solutions Chapter 9 Some Applications of Trigonometry Ex 9.1 23
⇒ h = ± 6 m
∴ h = 6 m [∵ Height is can be only positive]
Thus, the height of the tower is 6 m.

We hope the given KSEEB SSLC Class 10 Maths Solutions Chapter 12 Some Applications of Trigonometry Ex 12.1 will help you. If you have any query regarding Karnataka SSLC Class 10 Maths Solutions Chapter 12 Some Applications of Trigonometry Exercise 12.1, drop a comment below and we will get back to you at the earliest.

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