Trigonometry Class 10 Extra Questions Kerala State Syllabus Maths Chapter 6

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SSLC Maths Chapter 6 Trigonometry Important Questions and Answers

Trigonometry Class 10 Extra Questions Kerala Syllabus

Trigonometry Class 10 Kerala Syllabus Extra Questions

Properties of 45°, 45°, 90° triangles and 30°, 60°, 90° triangles.

Question 1.
A bridge of length 600 meter is built across a river. It makes 45° with the stream. The width of the river is

(a) 300√2 m
(b) 200 m
(c) 300√3 m
(d) 100√2m
Answer:
300√2 m

Question 2.
Area and perimeter of a square are numerically equal. What is the length of its diagonal?
(a) 10
(b) 12√2
(c) 2√2
(d) 4√2
Answer:
(d) 4√2

Question 3.
One side of a rhombus is 10cm and one angle 45°

(a) What is the distance between the parallel sides marked in the figure ?
(b) What is the area of the Rhombus?
Answer:
(a) 5√2 cm
(b) 10 × 5√2 = 50√2 cm²

Question 4.
Sides of a parallelogram are 12 and 10 centimetres. Angle between them is 45°

(a) What is the distance between the parallel sides marked in the figure ?
(b) What is the area of the parallelogram?
Answer:
(a) 5√2 cm
(b) 10 × 5√2 = 60√2 cm²

Question 5.
ABCD is a parallelogram in which two sides are 18cm and 12 cm. Angle between two sides is 60°

(a) What is distance between the parallel sides?
(b) What is the area of the parallelogram?
Answer:
(a) 6√3 cm
(b) 18 × 6√3 = 108√3 cm²

Question 6.
ABC is a right triangle in which ∠B = 90°, ∠C = 60° and AB = 12 cm

(a) What is the length of the side BC?
(b) What is the side AC?
(c) Find the area of the square ACDE
Answer:
(a) 4√3 cm
(b) 8√3cm
(c) 192 cm²

Question 7.
ABC is an equilateral triangle of side 12cm in which AD is perpendicular to BC.

(a) What are the angles of A ADB?
(b) What are the sides of triangle ADB
(c) What is the area of the square ADEF?
Answer:
(a) ∠A =30° , ∠B = 60° ∠D = 90°
(b) AB =12 cm, BD = 6 cm, AD = 6√3 cm
(c) Area = (6√3)² = 36 × 3 = 108 cm²

Question 8.
ABCD is a rhombus of side 8cm, ∠D = 150°

(a) What is ∠A
(b) What is the altitude from D to AB
(c) Find the area of the rhombus ?
Answer:
(a) 180 – 150 = 30°
(b) 4 cm
(c) 32 cm²

Question 9.
Look at the sequence given below. It is the sequence of equilateral triangles of side 2, 4, 6 ………….

(a) Write the sequence of the length of altitudes?
(b) Find the altitude of 10th triangle
(e) Write the area of its tenth triangle?
(d) Write the algebraic form of the sequence of altitudes.
Answer:
(a) √3, 2√3, 3√3
(b) 10√3
(c) 100√3
(d) x_n = n√3

Question 10.
What is the altitude of an equilateral triangle of side 6cm ?
(a) 3√3 cm
(b) 2√3 cm
(c) 5√3 cm
(d) √2 cm
Answer:
373 cm

Question 11.
Length of the diagonal of a rectangle is 12 cm and it makes 30° with the longest side. What is the breadth of the rectangle?
(a) 7
(b) 6
(c) 5
(d) 4
Answer:
6 cm

Question 12.
In the figure O is the centre of the angle ∠ACB =30°

(a) What kind of triangle is OAB?
(b) If the radius of the circle is 12cm then what is the altitude of triangle OAB1
Answer:
(a) Equilateral triangle.
(b) 6√3 cm

Question 13.
In the figure ∠A = 60°, BC = 10 cm

(a) Draw the diameter BD in the figure and join DC
(b) What is the radius of the circle?
Answer:
(a) Draw yourself
(b) \(\frac{10}{\sqrt{3}}\) cm

Question 14.
In triangle ABC, the line AD is perpendicular to BC, AB = 12 cm

(a) What is the length of AD?
(b) What is the length of AC?
(c) What is the length of BC?
(d) Calculate the area of triangle ABC?
Answer:
(a) 6√3 cm
(b) 6√6 cm
(c) 6 + 6√3 cm
(d) 54 + 18√3 cm²

Question 15.
The diagonal of the rectangle ABC is 12cm, ∠BAC – 30°

(a) What is the length of the side AB?
(b) What is the length of the side BC?
(c) Calculate the area of the rectangle
Answer:
(a) 6√3 cm
(b) 6 cm
(c) 36√3 cm²

Question 16.
Which of the following is the smallest?
(a) sin 72°
(b) cos 2°
(c) sin 10°
(d) cos 89°
Answer:
cos 89°

Question 17.
In triangle ABC, angle ∠A = 60°, BC = 12cm . The radius of the circumcircle is
a) 4√3
b) 3√3
c) 5√3
d) 4√2
Answer:
4√3cm

Question 18.
In triangle ABC, ∠A = 60°, AB = 10, AC = 12

(a) What is the altitude CD?
(b) Find the area of triangle ABC?
Answer:
a) 6√3
b) \(\frac{1}{2}\) × 10 × 6√3 = 30√3 sq.units

Question 19.
In the figure ∠B =140°, AC = 12 cm

(a) What is the measure of ∠P?
(b) What is the radius of the circle?
[sin 40 = 0.64, cos40 = .76, tan 40 = .83]
Answer:
(a) 180 – 140 = 40°
(b) sin 40° = \(\frac{12}{AP}\)
0.64 = \(\frac{12}{AP}\), AP = 18.75, r = 9.31 cm

Question 20.
In the figure ∠A = 120°, AB = 6, AC = 10

a) What is the altitude from C to AB ?
b) Calculate the area of triangle ABC?
Answer:
a) ∠CAP = 180 – 120 = 60°
sin 60 = \(\frac{P}{10}\)
PC = 5√3

b) Area = \(\frac{1}{2}\) × 6 × 5√3
= 15 √3 sq.cm

Question 21.
In the figure a, b, c are the sides opposite to A, B and C, AP is the altitude to BC

a) Write AP in terms of c and sin B
b) Write the expression for the area of triangle ABC
c) If a = 12, c = 6 and ∠B = 60° then the area of
Answer:
a) sin B = \(\frac{AP}{c}\)
AP = c sin B

b) Area = \(\frac{1}{2}\) × BC × AP
A = \(\frac{1}{2}\) × a × c sin B sq.unit

c) Area = \(\frac{1}{2}\) × 12 × 6 sin60
= 36√3 Sq.units

Question 22.
In triangle ABC, a, b, c are the sides opposite to the angles A, B and C and R is the radius of the circumcircle.
a) Prove that \(\frac{b}{\sin B}\) = 2R ?
Answer:

Draw the diameter AP and join PC. Triangle APC is a right triangle ∠B = ∠P
sin B = \(\frac{1}{2}\)
where R is the radius of the circumcircle.

b) Establish the relation the area of the circle A = \(=\frac{a b c}{4 R}\)?
Answer:
We know that
Area = \(\frac{1}{2}\) × ac × sin B
= \(\frac{1}{2}\) ac × \(\frac{b}{2 R}=\frac{a b c}{4 R}\)

Question 23.
When the angle measure increases from 0 to 90° its sin measure increases from 0 to 1 The maximum value of sin measure is 1.
When the angle measure increases from 0 to 90° its cos measure decreases from 1 to 0.
The maximum value of cos measure is 1
If the sum of two acute angles is 90°then the sin of one angle is equal to cos of other angle.
a) If sin A = cos A then what is A ?
b) If sin A = cos B then what is A + B ?
c) What is cos 1°x cos 2°x………. cos 90°?
d) If sin A + sin B + sin C = 3 then what is cos A + cos B + cos C ?
Answer:
a) 45°
b) 90°
c) 0
d) 0

Question 24.
From a point on the ground 40 metre away from the foot of the tower sees the top of the tower at an angle of evation 30° and sees the top of the water tank on the top of the tower at an angle of elevation 45°
a) Draw a rough diagram.
Answer:

b) Find the height of the tower.
Answer:
In triangle ABD we have tan 45° = \(\frac{BD}{AB}\)
l = \(\frac{h+h_1}{40}\), h + h₁ = 40
tan 30° = \(\frac{B C}{A B}, \frac{1}{\sqrt{3}}=\frac{h}{40}\)
h = \(\frac{40}{\sqrt{3}}\) = 23.1 m
Height of the tower 23.1 m

c) Find the height of the water tank
Answer:
23.1 + h₁ =40, h₁ = 40 – 23.1 = 16.9meter.

Question 25.
A man standing on the bank of a river sees the top of the tree at an angle of elevation 50° .Stepping 20 m back finds the angle to sees at an angle of elevation 30°
a) Draw a rough diagram.
Answer:
Diagram

b) Find the width of the river.
Answer:
tan 50° = \(\frac{BC}{AB}\), tan 30° = \(\frac{BC}{BD}\)

1.19 × √3x = x + 20
1.19 × 1.73x = x + 20
x = 18.89 m
width of the river 18.89 metre.

c) Calculate the height of the tree.
Answer:
When x = 18.89, h = 1.19 x = 1.19 × 18.89
= 22.47 metre

Question 26.
A tree of height 12 m is broken by the wind . The top struck the ground at an angle of 60°.
a) Draw a rough diagram
Answer:
Diagram

b) At what height from the bottom of the tree is broken by the wind?
Answer:
sin 60° = \(\frac{AC}{DC}\)
\(\frac{\sqrt{3}}{2}=\frac{x}{12-x}\)
12√3 – √3x = 2x,
12√3 = (2 + √3)x
x = 5.56 meter

c) Calculate the distance from the foot of the tree to its tip on the on the ground.
Answer:
tan 60° = \(\frac{x}{DA}\)
√3 = \(\frac{x}{DA}\), 1.73 = \(\frac{5.56}{DA}\)
DA = \(\frac{5.56}{1.73}\) = 3.2 m

Question 27.
The shadow of a vertical tower on level ground increases by 10m when the altitude of the Sun changes from the angle of elevation 45° to 30°
a) Draw a rough diagram
Answer:
Diagram

b) Calculate the height of the tower.
Answer:
tan 45° = \(\frac{AB}{AC}\), l = \(\frac{h}{x}\), h = x
tan 30° = \(\frac{1}{\sqrt{3}}=\frac{h}{x+10}\)
x + 10 = √3h, h = 13.56 metre.

Question 28.
A tall building and a short building are standing on a level ground.The angle of elevation of the top of the short building from the foot of the tall building is 30°.
The angle of elevation of the top of the tall building from the foot of the short building is 60° . The tall building has height 50m.
a) Draw a rough diagram
Answer:
Diagram

b) What is the distance between the buildings.
Answer:
tan 60° = \(\frac{50}{x}\), √3 = \(\frac{50}{x}\)
x = \(\frac{50}{\sqrt{3}}=\frac{50}{1.73}\) = 28.9 m
The distance between the buildings = 28.9 m

c) Calculate the height of the short building.
Answer:
tan 30 = \(\frac{h}{x}\)
\(\frac{1}{\sqrt{3}}=\frac{h}{28.9}\)
√3h = 28.9
h = \(\frac{28.9}{1.73}\) = 16.7 m

Question 29.
The distance between two buildings is 100 metre.The height of one building is double the height of other building.The top of the buildings can be seen at the angle of elevations 60° and 30° from a point in between the buildings.
a) Draw a rough diagram
Answer:

b) What is the diatance from the foot of the tall tower to the point of observation.
Answer:
Let AC = h and BD = 2h
tan 30° = \(\frac{h}{100-x}\), 100 – x = √3h
tan 60° = \(\frac{2 h}{x}\), √3x = 2h,
h = \(\frac{\sqrt{3} x}{2}\)
100 – x = √3h,
100 – x = \(\frac{\sqrt{3} \times \sqrt{3} x}{2}=\frac{3 x}{2}\)
200 – 2x = 3x,
5x = 200,
x = 40 m
The point of observation is at the diatance 40 m from tall building.

c) h = \(\frac{\sqrt{3} x}{2}=\frac{\sqrt{3} \times 40}{2}\) = 20√3 m
Height of the small building is 20√3 m and height of the tall building is 40√3 m

Question 30.
The top of a 30 high building can be seen from a point at some distance from the foot of the building is at an angle of elevation 30°.When the point of observation is some distance closer to the building the angle become 60°.
a) Draw a rough diagram
Answer:

b) What is the distance from the foot of the building to the second point of observation.
Answer:
In ∆ABP, tan 60° = \(\frac{30}{y}\)
3 = \(\frac{30}{y}\)
y = \(\frac{30}{\sqrt{3}}\) = 10
The second point is at the distance 10 × 1.732 = 17.32 m away from the building.

c) What is the distance between two points from which the top of the building is observed .
Answer:
tan 30° = \(\frac{30}{x+y}\)
\(\frac{1}{\sqrt{3}}=\frac{30}{x+y}\) , 30√3 = x + y = x + 17.3,
x = 30√3 – 17.3 = 34.66 m
The distance between the points is 34.66 m

d) What is the distance from the foot of the tower to the first point of observation.
Answer:
Distance is x + y = 30√3 = 51.96 m

Question 31.
A baloon, moving horizontally at the height 88.2 m from the ground sees at an angle of elevation 60° from a point on the ground. After some time it can be seen at the angle 40° from the same point
a) Draw a rough diagram.
Answer:

b) Calculate the diatance moved by the baloon during this time.
Answer:
In ∆ABP, tan60° = \(\frac{88.2}{P A}\),
PA = \(\frac{88.2}{1.73}\) = 50.98 m 1.73
tan 40 = \(\frac{88.2}{PC}\), PC = \(\frac{88.2}{0.83}\) = 106.2 m
Distance moved by the baloon is 106.2 – 50.98 = 55.2 metre.