Trigonometry Questions and Answers Class 10 Maths Chapter 6 Kerala Syllabus Solutions

Students often refer to Kerala Syllabus 10th Standard Maths Textbook Solutions Chapter 6 Trigonometry Questions and Answers Notes Pdf to clear their doubts.

SSLC Maths Chapter 6 Trigonometry Questions and Answers

Trigonometry Class 10 Questions and Answers Kerala State Syllabus

SCERT Class 10 Maths Chapter 6 Trigonometry Solutions

Class 10 Maths Chapter 6 Kerala Syllabus – Angles And Sides

(Textbook Page No. 102)

Question 1.
Calculate the area of parallelograms shown below

Answer:
In the first diagram triangle AMD is a 45° – 45° – 90° triangle

Draw perpendicular to AB. Since AD = 2 cm , DM = \(\frac{2}{\sqrt{2}}=\) = √2 cm
Area of the parallelogram is 4 × √2 = 4√2 cm²
In the second diagram triangle AMD is a 30° – 60° – 90° triangle
AD = 2 cm , DM = √3 cm.
Area of the parallogram = 4 × √3 = 4√3 cm²

Question 2.
Calculate the area of triangles shown below:

Answer:
Draw CM perpendicular to AB in all diagrams

AMC is a 30° – 60° – 90° triangle
CM = 1 cm
Area of the triangle is = \(\frac{1}{2}\) × 3 × 1 = \(\frac{3}{2}\) cm²

In the second diagram AMC is a 30° – 60° – 90° triangle CM = √3 cm.
Area of the triangle is= \(\frac{1}{2}\) x 3 x √3 = \(\frac{3}{2}\)√3 cm²

In the third diagram AMC is a 45° – 45° – 90°
In third triangle, CM = √2
Area of the triangle is \(\frac{1}{2}\) × 3 × √2 = \(\frac{3}{2}\) √2 cm²

Class 10 Maths Kerala Syllabus Chapter 6 Solutions – New Measures Of Angles

(Textbook Page No. 110)

Question 1.
The lengths of two sides of a triangle are 8 centimetres and 10 centimetres and the angle between them 40°
(i) What is the area of this triangle?
(ii) What is the area of the triangle with lengths of two sides the same, but the angle between them 140°?
Answer:
(i) In the figure

sin 40° = \(\frac{h}{8}\), 0.6428 = \(\frac{h}{8}\), h = 5.14 cm
Area = \(\frac{1}{2}\) × 10 × 5.14
= 25.70 cm²

(ii) In the second figure
h = 5.14 and area is 25.70 cm²

Question 2.
The length of sides of a rhombus is 5 centimetres and one of its angles is 100°. Calculate its area.
Answer:
In the figure ∠A = 180 – 100 = 80°

sin 80° = \(\frac{h}{2}\), 0.9848 = \(\frac{h}{5}\) ⇒ h = 4.9240
Area = 5 × 4.9240
= 24.62 cm²

Question 3.
The lengths of the diagonals of a parallelogram are 8 centimetres and 12 centimetres and the angle between them 50°. Calculate its area.
Answer:

\(\frac{AM}{6}\) = sin 50° ⇒ AM = 6 × 0.76 = 4.56
Area of triangle ABD = \(\frac{1}{2}\) × 8 × 4.56 = 18.24 cm²
Area of parallelogram = 2 × 18.24 = 36.48 cm²

Question 4.
A triangle is to be drawn with one side 8 centimetres long and one of the angles on it 40°. What is the minimum length of the side opposite this angle?
Answer:
We know that shortest distance from a point to a line is the perpendicular distance
BC is the perpendicular distance. Triangle ACB is a right triangle

See the diagram

\(\frac{BC}{8}\) = sin 40 = 0.6427
⇒ BC = 8 × 0.6427 = 5.14cm
Minimum length of the side opposite to 40° angle is 5.14 cm

Question 5.
The length of the sides of a rhombus is 5 centimetres and one of its angles is 70°. Calculate the length of its diagonals.
Answer:
See the diagram

sin 35° = \(\frac{b}{5}\) ⇒ b = 5 × 0.5735 = 2.86 cm
cos 35° = \(\frac{a}{5}\) ⇒ a = 5 × 0.8191 = 4.09 cm
Length of diagonals = 2 × 2.86 = 5.72 cm ,
= 2 × 4.09 = 8.18 cm

SCERT Class 10 Maths Chapter 6 Solutions – Triangles And Circles

(Textbook Page No. 114)

Question 1.
The pictures below show two triangles and their circumcircles:

Calculate the radius of each circle.
Answer:
Draw a rough diagram and mark the measurements.

Draw the diameter BP and join PC Triangle
BPC is a right triangle. ∠P = 30°
By the property of 30° – 60° – 90° triangle, Diameter 4.
Radius 2 cm

Draw a diameter AP, join PC
APCB is a cyclic quadrilateral.
∠P= 180 – 135 = 45°
∆ACP is a 45°, 45°, 90°
∆AC = PC = 3cm
AP = 3√2
Radius of the circle \(\frac{3}{2}\)√2cm

Question 2.
A circle is to be drawn, passing through the ends of a line 5 centimetres long; and the angle in one of the segments made by the line should be 80°. What should be the radius of the circle?
Answer:
See the diagram

sin 80° = \(\frac{5}{2r}\), 0.9848 = \(\frac{5}{2r}\)
r = \(\frac{5}{2 \times 0.9848}\) = 2.53 cm

Question 3.
The picture shows part of a circle:

What is the radius of the circle?
Answer:
See the digram

In the diagram the circle is completed by drawing the alternate arc.
Angle made by the chord in the alternate arc is 180 – 140 = 40°
Now we can complete a right triangle.
sin 40° = \(\frac{8}{2r}\) where r is the radius of the circle.
0.64 = \(\frac{8}{2r}\) ⇒ r = 6.25 cm
Note that, if a chord makes an angle c at the centre then length of chord will be 2r sin (\(\frac{c}{2}\))

SSLC Maths Chapter 6 Questions and Answers – Ratio Of Sides

(Textbook Page No. 118)

Question 1.
Two triangles and their circumcircles are shown in the figure.

Use the sine table to calculate the diameters of the circles and the other two sides of the triangles correct to a millimetre
Answer:

The length of chord d which makes a° angle on the arc. It can be found that d = 2r sin a°
4 = 2r sin 85°
⇒ r = \(\frac{2}{\sin 85^{\circ}}\) ≈ 2.01cm sin 85°
In the second diagram, angle is 80°
r = \(\frac{2}{0.98}\) = 2.04 cm

Kerala Syllabus Class 10 Maths Chapter 6 Solutions – Another Measure

(Textbook Page No. 121,122)

Question 1.
One angle of a rhombus is 50° and the shorter diagonal is 6 centimetres. What is its area?
Answer:
In the diagram

Diagonals of a rhombus bisect perpendicularly. In triangle APB, ∠BAP = 25°.
tan 25° = \(\frac{3}{AP}\)
AP = 6.433 cm
Area of triangle ABD = \(\frac{1}{2}\) × 6 × 6.43
= 19.29 cm²
Area of rhombus is half the product of diagonals. You can follow this method to area.

Question 2.
A ladder leans against a wall with its foot 2 metres away from the wall. The angle between the ladder and the ground is 40°. How high is the top of the ladder from the ground?
Answer:

tan 40° = \(\frac{h}{2}\)
⇒ h = 2 × tan 40°
= 2 × 0.83
= 1.66 m

Question 3.
Three rectangles are to be cut along the diagonals to make triangles which are then rearranged to form a regular pentagon as shown below

The sides of the pentagon must be 30 centimetres. What should be the lengths of the sides of the rectangles?
Answer:

AQ, CQ are the sides of large rectangle. BP, AP are the sides of small rectangles.
sin 54° = \(\frac{AP}{30}\) ⇒ AP = 24.27, AC = 48.54 cm
cos54° = \(\frac{BP}{30}\) ⇒ BP = cos 54 × 30 = 17.63 cm
sin 18° = \(\frac{CQ}{48.54}\) ⇒ CO = 14.99 cm
cos 18° = \(\frac{AQ}{48.54}\) ⇒ AQ = 46.16cm

Question 4.
The perpendiculars in the picture below are drawn 1 centimetre apart:

Prove that their heights are in an arithmetic sequence. What is the common difference?
Answer:
In the first right triangle, let x be the base and h h
the height. \(\frac{h}{x}\) = tan 40° ⇒ h = x tan 40°
For the second right triangle height = (x + 1) tan 40°
For the third right triangle height = (x + 2) tan 40°
xtan 40°, x tan 40° + tan 40°,
x tan 40° + 2 tan 40° ………………..
is the resulting sequence. It is an arithmetic sequence.
Common difference is tan 40°

Question 5.
Calculate the area of a regular pentagon of sides 10 centimetres.
Answer:
\(\frac{AP}{10}\) = sin 54°
⇒ AP = 8.09, AC = 16.18 cm

\(\frac{BP}{10}\) = cos 54°
⇒ BP = 10 × cos 54° = 5.8 cm

Area of triangle ABC is
\(\frac{1}{2}\) × 16.18 × 5.8 = 46.922 cm²

Area of triangle AED is 46.922 cm²
cos 18° = \(\frac{AQ}{AC}\) ⇒ AQ = 15.388 cm

Area of triangle ACD is
\(\frac{1}{2}\) × 10 × 15.38 = 76.94 cm²
Area of pentagon = 46.92 + 46.92 + 76.94
= 170.78 cm²

Class 10 Maths Chapter 6 Kerala Syllabus – Heights And Distance

(Textbook Page No. 125)

Question 1.
When the sun is seen at an angle of elevation of 40°, the length of a tree’s shadow is 18 metres
(i) What is the height of the tree?
(ii) What would be the length of the shadow, when the sun is at an elevation of 80°
Answer:
Diagram

(i) tan 40° = \(\frac{h}{18}\) ⇒ h = 15.10 m
(ii) tan 80°= \(\frac{15.10}{x}\) ⇒ x = 2.66 m

Question 2.
From top of a building, a person sees the foot of a shop 30 metres away at an angle of depression 25°. What is the height of the building?
Answer:
Diagram

tan 25° = \(\frac{h}{30}\)
⇒ h = 30 × tan 25
⇒ h = 13.98 m

Question 3.
From the top of an electric post, two wires are stretched to either side and attached to the ground, making angles 55° and 45° with the ground. The distance between the feet of the wires is 25 metres. What is the height of the post?
Answer:
Diagram

tan 55° = \(\frac{x}{25-x}\)
⇒ 1.42 = \(\frac{x}{25 – x}\)
⇒ 35.5 = 2.4x
⇒ x = \(\frac{35.5}{2.4}\)
⇒ x = 14.79 m

Question 4.
A person, 1.75 metres tall, standing at the foot of a tower sees the top of a hill 40 metres away at an elevation of 60°. From the top of the tower, he sees it at an elevation of 50°. Calculate the heights of the tower and the hill.
Answer:

\(\frac{DE}{CE}\) = tan 50° ⇒ DF = 40 × tan 50° = 47.67 m
\(\frac{DF}{BE}\) = tan 60° ⇒ DF = 40 × tan 60° = 69.28 m
EF = 69.28 – 47.67
= 21.61m
Height of the tower AC = 21.61 + 1.75
= 23.36 m
Height of the hill DG = 47.67 + 23.36
= 71.03 m

Question 5.
A boy, 15 metres tall, standing at the edge of a canal sees the top of a tree on the other edge at an angle of elevation of 80°. Stepping 15 metres back, he saw it at an angle of elevation of 40°. How wide is the canal and how tall is the tree?
Answer:
See the diagram

\(\frac{y}{x}\) = tan 80°, \(\frac{y}{x+15}\) = tan40°
y = x tan80°, Therefore \(\frac{x \tan 80}{x+15}\) = tan 40
x = \(\frac{15 \tan 40^{\circ}}{\tan 80^{\circ}-\tan 40^{\circ}}=\frac{12.58}{4.8}\) = 2.62 m

Width of the canal = 2.62
Height of the tree y + 1.5 = x tan 80 +1.5
= 16.35m

Trigonometry Class 10 Notes Pdf

Class 10 Maths Chapter 6 Trigonometry Notes Kerala Syllabus

Introduction
In our high school class we introduce trigonometry as an extension of the similarity of triangles. Whenever a triangle is scaled its angles and ratio of sides remain unchanged. That means angle can be measured using ratio of sides. The measurement of angles using ratio of sides are called trigonometric measure of angles.

The unit starts with properties of two special triangles. First one is 45°- 45°- 90° triangles. Such a triangle can be obtained by drawing diagonal to a square. Second special triangle is 30° – 60° – 90° triangle. The altitude of a equilateral triangle divides the triangle into two such triangles.

The relation between sides and angles of these triangles give an insight into the trigonometric measures of the angle. Sine measure, Cosine measure and Tangent measure are three basic trigonometric measures of an angle.

Problems related to circles and triangles are solved in the unit with the help of trigonometry. Calculation of heights and distances are another area of discussion.
A trigonometric table is given at the end of the unit. This is helpful for solving problems which contain angles between 1° and 89°.

→ The diagonal divides a square into two right triangles.These triangles are equal and have angles 45°, 45°, 90°

→ On scaling a triangle angles remains same. The sides opposite to equal angles change proportion-ally.
If the sides opposite to 45° angles is a then the side opposite to 90° angle will be a √2a

→ Altitude of an equilateral triangle divides the triangle into two right triangles. These are equal triangles
Each right triangle have angles 30°, 60° and 90°
If the side of the equilateral triangle is 2, clearly it will be the side opposite to 90° angle of the right triangle
The side opposite to 30° angle is 1 , the half of the side opposite to 90° angle.
Side opposite to 60° angle is √3.

→ If the side opposite to 90° angle is ‘a ’ then the side opposite to 30° will be \(\frac{a}{2}\), side opposite to 60° is \(\frac{a}{2}\)√3

→ The sides of a right triangle change proportionally on scaling the triangle. For an acute angle a° of the right triangle the ratio of opposite side to the hypotenuse is the sine measure of that angle.
The ratio of adjacent side to the hypotenuse is the cosine measure of that angle .

→ In the figure

sin a° = \(\frac{p}{d}\), cos a° = \(\frac{q}{d}\)

→ Note that, in this proportion sin a° is the proporionality constant in the change of height of the perpendicular with respect to the distance of the point from the vertex.
Also, cos a° is the proportionality constant in the change of distance from the vertex of the angle to the foot of perpendicular with respect to the distance of the point from the vertex.

→ Note that, sin 30° = cos 60° = \(\frac{1}{2}\)
sin 60° = cos 30° = \(\frac{\sqrt{3}}{2}\)
sin 45° = cos 45° = \(\frac{1}{\sqrt{2}}\)

→ In a triangle, p and q he the two sides and a0 is the angle between these sides then area of the triangle is \(\frac{1}{2}\)pq sin a°

→ The length of an arc of a circle can be calculated using the central angle of the arc.
Apart from degree measure Radian is another measure of angle. This related to central angle of the arc of a circle and its radius

If s is the arc length and r is the radius then \(\frac{s}{r}\) will be the central angle of the arc in radian measure

→ The length of any chord of a circle is twice the product of the radius and the sine of half the central angle

→ Consider a point on one side of an angle of a0,and the perpendicular from this point to the other side; if the height of the perpendicular is p and the dis-tance of the foot of the perpendicular from the vertex is q, then.
tan a° = \(\frac{p}{q}\)

→ Trigonometric measures of angles ranging from 1° to 89° is given in the table.
When angle increases from 0 to 90° sin measure increases from 0 to 1 and cos measure decreases from 1 to 0
If the sum of two angles is 90° then sine of one angle is equal to cosine of other angle.
sin 1° = cos 89°, sin 10° = cos 80°, sin 45° = cos 45°

→ In any triangle with angles 45°, 45°, 90° both the shorter skies have the same length and the length of the longest side is √2 times this length

→ In any triangle with angles 30°, 60°, 90° the length of the longest side is 2 times the length of the shortest side; and the length of the side of medium size is √3 times the length of the shortest

→ In an isosceles triangle the perpendicular from the top vertex to the bottom side bisects the top angle.

→ Consider a point on one side of an angle of o°, at distance d from the vertex, and the perpendicular from this point to the other side; if the height of the perpendicular is p and the the distance of the foot of the perpendicular from the vertex is q, then.

sin a° = \(\frac{p}{d}\)
cos a° = \(\frac{q}{d}\)

→ The length of any chord of a circle is twice the product of the radius and the sine of half the central angle

→ The central angle of each chord is twice the angle opposite to it in the triangle.

→ Consider a point on one side of an angle of a°, and the perpendicular from this point to the other side; if the height of the perpendicular is p and the distance of the foot of the perpendicular from the vertex is q, then.

tan a° = \(\frac{p}{q}\)

→ Line of sight: It’s the imaginary line drawn from the eye of an observer to the object being viewed.

→ Angle of Elevation: The angle formed upwards between the horizontal line and the line of sight, when the observer looks up at an object.

→ Angle of Depression: The angle formed downwards between the horizontal line and the line of sight, when the observer looks down at an object.