Kerala Syllabus 10th Standard Maths Solutions Chapter 5 Trigonometry

Kerala Syllabus 10th Standard Maths Solutions Chapter 5 Trigonometry

Kerala State Syllabus 10th Standard Maths Solutions Chapter 5 Trigonometry

Trigonometry Text Book Questions and Answers

Textbook Page No. 103

Trigonometry Class 10 Kerala Syllabus Question 1.
In the triangle shown, what is the perpendicular distance from the top vertex to the bottom side? What is the area of the triangle?

Answer:

The sides of the right angle triangle with angles 30°, 60°, 90° are proportional to the number 1 : √3 : 2
AD : BD : AB = 1 : √3 : 2
AB = 2x = 4cm
x = 2 cm
AD = 2 cm
BD = 2√3cm
Perpendicular distance from the top vertex to the bottom side = AD = 2 cm
BC = 2BD = 4√3 cm
Area of triangle = 1/2 bh
1/2 × 4√3 × 2 = 4√3 cm²

Sslc Trigonometry Solutions Kerala Syllabus Question 2.
In each of the following parallelograms, find the distance between the top and bottom side? Calculate the area of parallelogram.

Answer:

Sslc Maths Chapter 5 Kerala Syllabus Question 3.
A rectangular board is to be cut along the diagonal and the pieces rearranged to form an equilateral triangle as shown below. The sides of the triangle must be 50 centimetres. What should be the length and breadth of the rectangle?

Answer:
The sides of the right angle triangle with angles 30°, 60°, 90° are proportional to the number 1 :√3: 2

Sslc Maths Trigonometry Kerala Syllabus Question 4.
Two rectangles are cut along the diagonal and the triangles got are to be joined to an-other rectangle to make a regular hexagon as shown below:

If the sides of the hexagon are to be 30 centimetres, what would be the length and breadth of the rectangles?
Answer:

The sides of the right angle triangle APF with angles 30°, 60°, 90° are proportional to the number 1 :√3: 2

Length of the smaller rectangle = 25.98 cm
Breadth of the smaller rectangle = 15 cm
Length of the bigger rectangle = 51.96 cm
Breadth of the bigger rectangle = 30 cm

Trigonometry Sslc Kerala Syllabus Question 5.
Calculate the area of the triangle shown.

Answer:
Angles of the first triangle are in ratio 45 : 45: 90. So sides are in ratio x : x: √2 x
Angles of the second triangle are in ratio 30 : 60: 90, sides are in ratio, y :√3y: 2y

Textbook Page No. 109

Trigonometry Problems For Class 10 State Syllabus Question 1.
The lengths of two sides of a triangle are 8 centimetres and 10 centimetres and the angle between them is 40°. Calculate its area. What is the area of the triangle with sides of the same length, but angle between them 140°?
Answer:

If the angle between the sides is 140° sin 140 = sin(180 – 40) = sin 40 The area of the triangles will be same.

Trigonometry Questions For Class 10 Kerala Syllabus Question 2.
The sides of a rhombus are 5 centimetres long and one of its angles is 100°, Compute its area.
Answer:

Sslc Maths Trigonometry Notes Kerala Syllabus Question 3.
The sides of a parallelogram are 8 centimetres and 12 centimetres and the angle between them is 50°. Calculate its area.
Answer:

Trigonometry Class 10 Scert Kerala Syllabus Question 4.
Angles of 50° and 65″ are drawn at the ends of a 5 centimetres long line to make a triangle. Calculate its area.
Answer:
If diameter is d

Sslc Trigonometry Kerala Syllabus Question 5.
A triangle is to be drawn with one side 8 centimetres and an angle on it 40°. What should be the minimum length of the side opposite this angle?
Trigonometry Questions and Answer: The side opposite to 40° will be at least

Textbook Page No. 114

Class 10 Maths Chapter 5 Kerala Syllabus Question 1.
The figure shows a triangle and its circumcircle: What is the radius of the circle?

Answer:
∠BAC = 60°
∠BOC = 120°
The angle made by any arc of a circle on the alternate arc is half the angle made at the centre.

The sides of the right angle triangle ΔCOD with angles 30°, 60°, 90° are proportional to the number 1 : √3: 2 .

Sslc Maths Chapter 5 Solutions Kerala Syllabus Question 2.
What is the circumradius of an equilateral triangle of sides 8 centimetres?
Answer:
The sides of the right angle triangle OBD with angles 30°, 60°, 90° are proportional to

Hsslive Trigonometry Kerala Syllabus Question 3.
The figure shows a triangle and its circum.

i. Computer the diameter of the circle.
ii. Compute the lengths of the other two sides of the triangle.
Answer:

Maths Chapter 5 Class 10 Kerala Syllabus Question 4.
A circle is to be drawn, passing through the ends of a line, 5 centimetres long; and the angle on the circle on one side of the line should be 80°. What should be the radius of the circle?
Answer:

Maths Trigonometry Class 10 State Syllabus Question 5.
The picture below shows part of a circle:

What is the radius of the circle?
Answer:

First, complete the circle. Draw BD and join one of its end D to C. ∠BAC + ∠BDC= 180° ∠D = 40°, ∠BCD (angle on semicircle). So ABCD is right-angled.

Trigonometry Questions For Class 10 Scert Question 6.
A regular pentagon is drawn with all its vertices on a circle of radius 15 centimetres. Calculate the length of the sides of this pentagon.
Answer:
Sum of angles of pentagon
= (n – 2)180
= (5 – 2)180
= 540°
One angle of regular pentagon = 540/5 = 180°

Textbook Page No. 117

Kerala Syllabus 10th Standard Maths Chapter 5 Question 1.
One angle of a rhombus is 50° and the larger diagonal is 5 centimetres. What is its area?
Answer:
In rhombus ABCD, One angle of a rhombus is 50° and one diagonal is 5 centimetres.

Sslc Maths Chapter 5 Trigonometry Kerala Syllabus Question 2.
A ladder leans against a wall, with its foot 2 metres away from the wall and the angle with the floor 40°. How high is the top end of the ladder from the ground?
Answer:
tan 40 = QR/2
QR = 2 × tan 40
= 2 × 0.8391 = 1.6782
Height of the ladder from ground = 1.68m

Hss Live Guru 10th Maths Kerala Syllabus Question 3.
Three rectangles are to be cut along the diagonals and the triangles so got rearranged to form a regular pentagon, as shown in the picture. If the sides of the pentagon are to be 30 centimetres, what should be the length and breadth of the rectangles?

Answer:

36°, 54°, 90°
Sin 54 = DG/30
DG = 30 × 0.8090
= 24.27 cm
Cos 54° = EG/30
EG = 30 × 0.5878 = 17.63 cm
Length of larger rectangle = 46.17 cm.
Breadth = 15 cm
Length of smaller rectangle = 24.27 cm
Breadth = 17.63 cm

10th Standard Maths Trigonometry Kerala Syllabus Question 4.
In the picture, the vertical lines are equally spaced. Prove that their heights are in arithmetic sequence. What is the common difference?

Answer:

10th Trigonometry Questions And Answers Kerala Syllabus Question 5.
One side of a triangle is 6 centimetres and the angles at its ends are 40° and 65°. Calculate its area.
Answer:
∠C = 180 – (40 + 65) = 75°
Draw a perpendicular BD from B to AC

Textbook Page No. 122

Trigonometry Hsslive Kerala Syllabus Question 1.
When the sun is at an elevation of 40°, the length of the shadow of a tree is 18 metres. What is the height of the tree?
Answer:
In the right triangle ΔPQR
tan 40° = QR/18
QR = 18 × tan40° =18 × 0.8391
Height of the tree = 15.1 m

Class 10 Maths Chapter 5 Trigonometry Kerala Syllabus Question 2.
When the sun is at an elevation of 35°, the shadow of a tree is 10 metres. What would be the length of the shadow of the same tree, when the sun is at an elevation of 25°?
Answer:

Question 3.
From the top of an electric post, two wires are stretched to either side and fixed to the ground, 25 metres apart. The wires make angles 55° and 40° with the ground. What is the height of the post?
Answer:

Question 4.
A 1.5-metre tall boy saw the top of a building under construction at an elevation of 30°. The completed building was 10 metres higher and the boy saw its top at an elevation of 60° from the same spot. What is the height of the building?
Answer:
In the right triangle BDE,

Question 5.
A 1.75-metre tall man, standing at the foot of a tower, sees the top of a hill 40 metres away at an elevation of 60°. Climbing to the top of the tower, he sees it at an elevation of 50°. Calculate the heights of the tower and the hill.
Answer:
In the right triangle CEF
BD = CE = AF = HG = 40m

AB = tower
DF = hill
Height of hill = 69.28 + 1.75 = 71.03 m
In the right triangle HGF,
tan50=GF/HG=GF/40
GF = 40 × tan 50= 40 × 1.1918 = 47.67 m
Height of tower = 71.03 – (47.67 + 1.75)
= 71.03 – 49.42 = 21.61 m

Question 6.
A man 1.8 metre tall standing at the top of a telephone tower, saw the top of a 10-metre high building at a depression of 40° and the base of the building at a depression of 60°. What is the height of the tower? How far is it from the building?
Answer:

Height of the building = 10m
Height of the towerAG
Height of the man GF = 1.8m
AB = x
In the right triangle CHF
tan 40 = HF/x
HF = x tan 40
= x × 0.8391
= 0.8391x
In the right ABF
tan60=AF/AB=AF/x⇒
AF = x tan 60 = 1.732x
BC = AH = AF – HF
= 1.732x – 0.8391x = 10
= o.8929x = 10

Height of tower = 19.4 – 1.8 =17.6 m
Distance from building to tower = 11.2m

Trigonometry Orukkam Questions and Answers

Worksheet 1

Question 1.
Complete the table given below.

Answer:
∠C = 90°
The sides of the right angle triangle with angles 30°, 60°, 90° are proportional
1 : √3: 2

Question 2.
Complete the table given below.


Answer:
The sides of the isosceles right-angle triangle with angles 45: 45 :90 will have sides proportional to [opposite to corresponding angles] 1: 1: √2

Question 3.
Calculate the perimeter of the triangle.

Answer:
i. BC = 10
∴ AB = 10

Question 4.
ABC Dis a square AC = 10cm . Find B, BACFind the length of AB. Find the perimeter of the square.
Answer:
∠B = 90°
∠BAC = 45° = ∠BCA
AB=10/√2=5√2
Perimeter of the square D
= 4 × 5√2 = 28.28 cm

Question 5.
PQRS is a rectangle. Find angle SP R? Find angle PRQ. If PR = 30 then find P Qand QR. Calculate the perimeter of the rectangle.

Answer:

∠SPQ = 90° ( ∴ PQRS is a square, so angles are 90° each.)
∴ ∠SPR = 90 – 30 = 60°

Question 6.
If C D = 5 then find ∠ACD,∠BCD . Find AB, AD, BD, BC. Find the angles of triangle ABC. If the angles are 45°, 60°, 75° find the ratio of the sides?

Answer:

Worksheet 2

Question 7.
In the figure BC = 12 ∠D = 90°,Find ∠CBD, ∠ACD, ∠ABC . Find BD, C D, AD, AC, AB. Find the ratio of the sides of the triangle having the angles 30°, 15°, 135° C

Answer;
∠ACD = 180 – (30 + 90) = 60
∠BCD = 60 – 15 = 45 ∠CBD = 45
∠ABC = 135 [∵ 180 – (30 + 15)]

Question 8.
In the figure AD = 7, CD = 8, BD = 5, ∠ADP = 50° then find ADB ?

Answer:

Question 9.
Find the measure of the remaining part of the triangle from the figure given below

Question 10.
∠AOB = 2x, radius of the circle R. Find ∠AOC? Find sin x, AC and AB
Answer:

Worksheet 3

Question 11.
Using the figure find AB

Question 12.
In the figure BD = 10,

Find ∠BAD and ∠ADB.
AD, CD and AC
Answer:
∠ADB = 180 – 60 = 120°
∠BAD = 180 – (120 + 30) = 30

The sides of the right angle triangle with angles 30°, 60°, 90° are proportional to
1 :√3: 2
AD = 2 × 5√3 = 10√3
AC = 5 √3 × √3 = 15

Question 13.
In the figure QR = 7, find ∠QRP, ∠QPR Find the length of PR, PS and RS.

Answer:


Question 14.
In the figure BD = 10, CD = x, find the length of BC. Using tan 40, tan 50 find the length of AC.
Answer:

Worksheet 4

Question 15.
In triangle ABC , AB = 7, BC = 12, ∠B = 40 Find the area of the triangle. Calculate the length of AC.
Answer:
Draw a perpendicular line
AD from A to BC.
AD = AB sin40
= 7 × 0.6428 = 4.4996 = 4.5
Area = 1/2 × BC × AD
= 1/2 × 12 × 4.5 = 27m2
BD = AB cos 40 = 7 x 0.7660 = 5.36
∴ CD = 12 – 5.36 = 6.64

From right angled ΔADC

∴ Length of AC = 8.02 units

Question 16.
In triangle ABC , AB = 7, BC = 12, ∠B = 40 Find the area of the triangle.
Answer:
See the above question

Question 17.
In the figure AD = BD = C D = 5 ∠ADC = 50° . find the area of triangle AC D, triangle ABD and triangle ABC.

Answer:
AD = BD = CD, So ΔABD and ΔACD are isosceles triangles. AE is the perpendicular from D to AC

Area of ΔABD = AF × DF
= 4.5 × 2.1 = 9.45
Area of ΔABC= 1/2 × AB × AC
1/2 × 9 × 4.2 = 18.9m²

Question 18.
ABC D is a parallelogram, angle D = 120°, AB = 10, AC = 12. Calculate the area of the parallelogram.

Answer:
Area of the parallelogram= 1/2 × AC × BD
In ΔABD, ∠A = 60°
ΔABD is an equilateral triangle.
BD = 10
Area of the parallelogram ABCD = 1/2 × 10 × 12 = 60m²

Question 19.
One angle of a triangle is 30°, prove that radius of the circumcircle is equal to the side opposite to 30°
Answer:
For a right-angled triangle one of the angles is 30° then other one is 60°.
Side which is opposite to the angle of 90° is twice of the side which is opposite to the angle of 30°
Center of circumcircle is the midpoint of the side, which is opposite to the angle 90° that means half.
∴ Radius of the circumcircle is equal to the side opposite to 30°.

Question 20.
O is the centre of a circle having a chord
AB. AB= 12, angle AOB = 120°. Find the radius
Answer:
AC = BC = 6
A perpendicular is drawn through center which can bisect the perpendicular line AB into half.
In ΔAOC , ∠AOC = 60°, ∠ACO = 90°


Question 21.
Above viewed the top of a tree at an angle of elevation 30°. He moved 10 m towards the tree and saw the top of the tree ant the angle 60° Find the height of the tree
Answer:

Question 22.
In the figure BC =14, ∠5 = 40°, ∠C = 50° Find the area of triangle ABC.
Answer:

Area of ΔABC =1/2 × AB × AC
AC = BC sin 40 = 14 × 0.6428 = 8.99
AB = BC cos 50 = 14 × 0.7660 = 10.72
Area of ΔABC = 1/2 × 8.99 × 10.72 = 48.2 m²

Worksheet 5

Question 23.
A child observed an aeroplane flying horizontally at the height 1km at ah angle of elevation 60°at an instant. After ten seconds he saw the plane at the angle 30°. Calculate the speed of the plane.
Answer:

speed = distance/time = 1150m/ 10sec = 115 m/s

Question 24.
In the figure BC = a, CD = b. Then prove that a = 3b.
Answer:

Worksheet 6

Question 25.
A man observed the top of a tower at a distance a from its base at an angle of elevation 60°. He saw the top of the tower at an angle of elevation 30° from a point at the distance b from the base. Prove that height of the tower h = √ab
Answer:

Trigonometry SCERT Questions and Answers

Question 26.
The diagonal of a square is 4cm long. Find its perimeter and area. [Score: 2, Time: 3 minute]

Question 27.
AC and BC are two equal chords of a circle with diameter AB. If the equal chords have lengths 10cm find the area of the circle. [Score: 3, Time: 3 minute]

10 Diameter AB = 10√2 cm. (1)
Radius = 5√2 cm (1)
Area = πr² = π × (5√2)² = 5π sq. (1)

Question 28.
In ΔABC, AB = 10 cm. AC 8 cm, ∠A = 45°
a. Find the perpendicular distance from C to AB.
b. Find the area of the triangle. [Score: 2, Time: 3 minute]

Answer:

Question 29.
One side of a rhombus is 12 cm and one angle is 135°.
a. Find the distance between the parallel sides?
b. Find the area of the rhombus. [Score:3,Time:3minute]

Answer:
a. ABCD is a rhombus
AB = AD = 12 cm, ∠B = 135° ∠A = 180 – 135 = 45°
Angles of ΔADE are 45°, 45°, 90°

Question 30.
In ΔABC ∠A = 45°, BC = 6 c m. Find the diameter of the circumcircle. [Score: 4, Time:4 minute]

Answer:
Draw diameter BD and Join CD. Angles of ΔBCD is 45°, 45°, 90° (1)

Question 31.
12 centimetre long diagonal of a rectangle makes an angle 30° With its one side. Find its perimeter and area. [Score: 4, Time:4 minute]

Answer:
ABCD is a square
AC = 12cm , ∠BAC = 30° Angles of ΔABC are 30°, 60°, 90°

Question 32.
In ΔABC, AB = 20 cm ∠A = 30°, AC = 12 cm
a. Find the length of the perpendicular from C to AB.
b. Find the area of the triangle. [Score: 3, Time: 4 minute]

Answer:
a. Draw CD perpendicular to AB. Angles of ΔADC are 30°, 60°, 90° (1)

Question 33.
In ΔABC, AB = 8 cm , BC = 10 cm and ∠B = 60°
a. Find the area of the triangle ΔABC.
b. Find AC. [Score: 4, Time:6 minute]

Answer:

Question 34.
One angle of a triangle is 150° and its opposite side 3 centimetre. Find the diameter of its circumcircle. [Score: 3, Time:5 minute]

Answer:
In DABC, ∠B = 150°, AC = 3 cm
Draw diameter AD and join CD (1)
∠ADC = 180 – 150 = 30°, ∠ACD = 90°
Angles of DADC are 30°, 60°, 90°
30° 60° 90°
1 : √3 : 2
↓    ↓     ↓
3 3√3 6 (1)

Diameter, AD = 6 cm (1)

Question 35.
In ΔABC AB = 12 cm, ∠A=45° and ∠B = 30°
a. Find the area of the triangle ABC.
b. Find the ratio of the sides of the triangle having angles 30°,45°,105° [Score: 5, Time:8 minute]

Answer:
a. Angles of ΔABC are 45°, 45°, 90°
Angles of ΔBDC are 30°, 60°, 90°. (1)

Question 36.
The diagonal of a rectangle is 12 cm and it makes an angle 35° With one side. Find the perimeter of the rectangle. [sin 35° = 0.57, cos 35° = 0.82] [Score, 3, Time: 5 minute]

Answer:
Let the breadth of the rectangle = x and length = y
sin 35° = x/12 (1)
x = 12 × sin 35 = 12 × 0.57 = 6.84 cm.
cos 35° = y/12
y = 12 × cos 35 = 12 × 0.82 = 9.84 cm. (1)
Perimeter = 2(6.84 + 9.84) = 2 × 16.68 = 33.36 cm (1)

Question 37.
In ΔABC, ∠A = 125°,BC = 8 cm. Find the diameter of the circumcircle. [sin 55 = 82] [Score: 3, Time:6 minute]

Answer:
Draw diameter BD and jon CD. (1)
In ΔBCD, sin55° = BC/BD (1)

Question 38.
Can one cut out a triangle of one side 7 cm and its opposite angle 40° from a circular sheet of diameter 10 cm. Justify your answer. [sin 40° = 0.64]
[Score: 4, Time:7 minute]
Answer:
The diameter of the circumcircle of a triangle with one angle 40° and it’s opposite
side 7 cm = 7/sin40 (1)
7/0.64 = 10.93 cm
Diameter of the paper is 10 cm, which is less than 10.93 cm.
Hence triangle cannot be cutout. (2)

Question 39.
Find the area of a triangle Whose sides are a and b and the angle between those sides is C. [Score: 2,Time:3 minute]
Answer:
In ΔABC Draw AD perpendicular to BC.
sin C = h/b
h = b sin C
Area = 1/2 ah
= 1/2 ab sin C (1)

Question 40.
Find the sides of a triangle whose angles are A, B and C and its circumdiameter d. [Score: 3, Time: 5 minute]
Answer:
Draw a diameter BD and join CD

∠BDC = A : BC = a (1)
sm A = a/BD = a/d
Similarly
b = d sin B° = AC (1)
AB = c = d sin C°. (1)

Trigonometry Exam Oriented Questions and Answers

Short Answer Type Questions (Score 2)

Question 41.
The angle of a right triangle is 30° and its hypotenuse is 4 cm. What is its area?

Answer:
Triangle side ratio is 1 : : 2
Altitude = hypotenuse ×√3/2=4×√3/2
Area = 1/2 × 23.46 = 3.46 cm²

Question 42.
In the figure, find the length of the side represented by x.

Answer:
tan50∘= opposite side / adjacent side =x/24
x = 2.4 × tan 50° = 2.4 × 1.1918 = 2.86

Question 43.
Calculate the area of a right-angled triangle whose one angle is 45° and hypotenuse 20 cm.
Answer:
Angle of the right 45°, 45°, 90°
Ratio of sides = 1 : 1 : √2
hypotenuse = 20cm
∴ The other two sides are 20/√2 cm each
∴ Area of the right angled triangle
=1/2×20/√2×20/√2=1/2×20×20/2=100cm2

Question 44.
Different sizes of isosceles triangle are given. In the table given below some of its sides are given. Fill the table.

Answer:
a. 4. 4, √32
b. 2, 2, √8
c. 3, 3, √18
d. 10, 10, √200
e. 1, 1, √2

Question 45.
The area of a parallelogram with one side 8cm and an angle 30° is 80 cm².
Find out the length of the other side.

Answer:

Area of the parallelogram = 80cm2
8h = 80
h = 80/8= 10cm
The angles of ΔAPD are 30°, 60° and 90°
The sides are in the ratio 1 : √3 : 2
DP = 10cm
∴ AD = 20cm

Short Answer Type Questions (Score 3)

Question 46.
In the figure, ∠BAC = 90°, AD = 6cm, CD = 9cm, ∠ACD = x.
a. What is tan x?
b. How much is ∠BA D
c. What is the length of BD?

Answer:

Question 47.
In the quadrilateral ABCD shown below,
∠A = ∠C = 90% ∠ABD = 45° ∠CDB = 60° and AB = 6cm.
Find out the lengths of the other sides of the quadrilateral.

Answer:
In ΔABD,
The angles are 45°, 45°, 90°
As the triangle is equilateral, sides are in the ratio, 1: 1: √2
AB = 6cm
∴AD = 6cm, BD = 6√2cm

Question 48.
P be a point on a circle having diameter AB. ∠ABP = 30°, BP = 6cm. Draw a rough figure.
Find the length of AP and area of circle ?
Answer:
∠P = 90°, ∠A = 60°,
AP: PB: AB = 1: √3 :2
AP = 2√3 cm,
AB = 4√3 cm
Radius of circle = 2√3 cm
Area of circle = π(2√3)2
cm = 12 π cm²

Long Answer Type ? Questions (Score 4)

Question 49.
In ΔABC, ∠A = 110° and BC = 8cm Find out the radius of the circumcircle.

Answer:
Draw diameter BD and Join D and C. The opposite angles of cyclic quadrilaterals are supplementary ∠D = 70°, BCD is a semicircle. ∠BCD is the angle in a semicircle, ∠BCD =90°

Question 50.
Length of two sides of a triangle are 20cm and 16cm and the angle between them is 135°.
a. Draw a rough figure and mark the measurements.
b. Find the perpendicular distance of the vertices to the side of length 20cm.
c. Find the area of the triangle.
Answer:

b. Now AADB is a right triangle with angles 45°, 45°, 90°. Since the side opposite to 90° angle is 16cm, the other two sides are 8√2 each.

Perpendicular distance of the vertex to the side of length 20cm is 8√2 cm.

Long Answer Type Questions (Score 5)

Question 51.
A girl standing on a lighthouse built on a cliff near the seashore, observes two boats due East of the lighthouse. The angles of depression of the two boats are 300 and 600. The distance between the boats is 300m.
a. Draw a rough figure based on the given details.
b. Find the distance of the top of the lighthouse from the sea level. (Boats and foot of the lighthouse are in a straight line).
Answer:

Distance of the top of the lighthouse from the sea level = 259.8m

Question 52.
In the figure, OR is perpendicular to OP and OP = 12cm. A, B and C are points on OR. If ∠OPA = 30°? ∠APB = 15°,and ∠BPC = 15°. Find OA, OB and OC. Also find AB: BC.

Answer:
ΔOPA is a right triangle with angles 30°, 60°, 90°. Since the side opposite to 30° angle is 4√3


Now consider ΔOPB. It is a right triangle with angles 45°, 45°, 9Q° Since the side OP = 12cm, side OB is also 12cm.

Also ΔOPC is a right triangle with angles 30°, 60°, 90°. Since the side opposite to 30° angle is 12cm, the side opposite to the 60° angle ie OC is 12√3.
Thus OA = 4√3 cm,

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