Mathematics Part I Solutions Solutions for Class 10 Maths Chapter 4 Trigonometry are provided here with simple step-by-step explanations. These solutions for Trigonometry are extremely popular among Class 10 students for Maths Trigonometry Solutions come handy for quickly completing your homework and preparing for exams. All questions and answers from the Mathematics Part I Solutions Book of Class 10 Maths Chapter 4 are provided here for you for free. You will also love the ad-free experience on Meritnationâ€™s Mathematics Part I Solutions Solutions. All Mathematics Part I Solutions Solutions for class Class 10 Maths are prepared by experts and are 100% accurate.

#### Question 1:

The area of a parallelogram is 30 square centimetres. One of its sides is 6 centimetres and one of its angles is 60°. What is the length of its other side?

Let ABCD be the given parallelogram.

DC = 6 cm and ADC = 60°

Area of parallelogram ABCD = 30 cm2

Construction: Draw AE perpendicular to DC.

Area of a parallelogram = Base × Height

30 cm2 = 6 cm × AE

AE =

Using angle sum property in ΔAED:

60° + 90° + EAD = 180°

⇒ ∠EAD = 180° 150° = 30°

Thus, the angles of ΔAED are 30°, 60° and 90°.

Therefore, the ratio of the lengths of the sides of ΔAED is 1::2.

Thus, the length of other side of the parallelogram is cm.

Note: We can also draw the figure by taking ADE = 60° and AD = 6 cm.

#### Question 2:

The sides of an equilateral triangle are 4 centimetres long. What is the radius of its circumcircle?

We know that the median, perpendicular bisector and angle bisector of an equilateral triangle coincide and pass through the centre of the circle.

Consider the figure given below.

Here, ΔABC is an equilateral triangle with side 4 cm.

Each angle of an equilateral triangle is of measure 60°.

∴ ∠ABD = 60°

The circle with centre O in the given figure is the circumcircle for ΔABC.

AD is the perpendicular bisector of side BC.

Using angle sum property in ΔADB:

ABD + BDA + DAB = 180°

60° + 90° + DAB = 180°

150° + DAB = 180°

⇒ ∠DAB = 180° 150° = 30°

Thus, the angles of ΔADB are 30°, 60° and 90°.

Therefore, the ratio of the lengths of the sides of ΔADB is 1::2.

We know that centroid of a triangle divides the median in the ratio of 2:1.

Thus, the radius of the circumcircle for the given triangle is .

#### Question 3:

One angle of a right angled triangle is 30° and its hypotenuse is 4 centimetres. What is its area?

Let ΔABC be the given right-angled triangle, right-angled at B.

Also, ACB = 30° and Hypotenuse = AC = 4 cm

Using angle sum property in ΔABC

ABC + BCA + CAB = 180°

90° + 30° + CAB = 180°

120° + CAB = 180°

⇒ ∠CAB = 180° 120° = 60°

Thus, the angles of ΔABC are 30°, 60° and 90°.

Therefore, the ratio of the lengths of the sides of ΔABC is 1::2.

#### Question 1:

In the figure below, O is the centre of the circle.

What is the diameter of the circle?

Given: O is the centre of the given circle.

As O is the centre of the circle, AC is the diametre of the given circle.

ABC is an angle in a semicircle.

We know that angle in a semicircle is a right angle.

∴ ∠ABC = 90°

Also, CAB = 30°

Using angle sum property in ΔABC:

ABC + BCA + CAB = 180°

90° + BCA + 30° = 180°

120° + BCA = 180°

⇒ ∠BCA = 180° 120° = 60°

Thus, the angles of ΔABC are 30°, 60° and 90°.

Therefore, the ratio of the lengths of the sides of ΔABC is 1::2.

AC = 2 × BC = 2 × 2.5 cm = 5 cm

Therefore, the diametre of the given circle is 5 cm.

#### Question 2:

What is the length of the chord in the figure below?

Construction: Draw OD perpendicular to AB.

In ΔAOB, OA = OB

⇒ ∠OBA = OAB (Angles opposite to equal sides are equal in measure.)

Using angle sum property in ΔAOB:

AOB + OBA + BAO = 180°

120° + 2BAO = 180°

2BAO = 180° 120°

2BAO = 60°

⇒ ∠BAO = 30°

Now, using angle sum property in ΔADO:

AOD + ODA + DAO = 180°

⇒ ∠AOD + 90° + 30° = 180°

⇒ ∠AOD + 120° = 180°

⇒ ∠AOD = 180° 120° = 60°

Thus, the angles of ΔADO are 30°, 60° and 90°.

Therefore, the ratio of the lengths of the sides of ΔADO is 1::2.

We know that perpendicular drawn from the centre to the chord bisects the chord.

Therefore, the length of the chord is

#### Question 3:

Two identical rectangles are to be cut along the diagonals and the triangles got joined with another rectangle, to make a regular hexagon as shown below:

What should be dimensions of the rectangles?

Given: ABHCDE is a regular hexagon made by using a rectangle and some pieces of rectangles cut along its diagonal.

We know that all the sides and angles of a regular hexagon are equal.

AB = BH = HC = CD = DE = EA = 30 cm

Measure of each angle of a regular hexagon = 120°

⇒ ∠EDC = DCH = CHB = HBA= BAE = AED = 120°

Measure of each angle of a rectangle = 90°

⇒ ∠ADC = EFD = 90°

Now, EDF = EDC ADC = 120° 90° = 30°

Using angle sum property in ΔEFD:

EFD + FDE + DEF = 180°

90° + 30° + DEF = 180°

120° + DEF = 180°

⇒ ∠DEF = 180° 120° = 60°

Thus, the angles of ΔEFD are 30°, 60° and 90°.

Therefore, the ratio of the lengths of the sides of ΔEFD is 1::2.

DF =

AF = DF = (As both are the sides of identical rectangles.)

AD = AF + FD =

EF =

Therefore, the dimensions of the larger rectangle are 30 cm and

The dimensions of the smaller rectangle are 15 cm and

#### Question 1:

Compute the lengths of all sides of the quadrilateral below:

Using angle sum property in ΔABC:

ABC +BCA + CAB = 180°

90° + 30° + CAB = 180°

120° + CAB = 180°

⇒ ∠CAB = 180° 120° = 60°

Thus, the angles of ΔABC are 30°, 60° and 90°.

Therefore, the ratio of the lengths of the sides of ΔABC is 1::2.

Now, using angle sum property in ΔADC:

90° + DCA + 45° = 180°

135° + DCA = 180°

⇒ ∠DCA = 180° 135° = 45°

Thus, the angles of ΔADC are 45°, 45° and 90°.

Therefore, the ratio of the lengths of the sides of ΔADC is 1:1:

Therefore, the length of all the sides of the given quadrilateral are 2 cm, , and

#### Question 2:

Compute the area of the rectangle below:

Given: ABDE is a rectangle.

We know that the sum of angles of a linear pair is 180°.

∴ ∠ACB + ACD = 180°

⇒ ∠ACB + 120° = 180°

⇒ ∠ACB = 180° 120° = 60°

In ΔABC, ABC is a right angle (angle of a rectangle).

Using angle sum property:

ABC +BCA + CAB = 180°

90° + 60° + CAB = 180°

150° + CAB = 180°

⇒ ∠CAB = 180° 150° = 30°

Thus, the angles of ΔABC are 30°, 60° and 90°.

Therefore, the ratio of the lengths of the sides of ΔABC is 1::2.

Now, using angle sum property in ΔABD:

ABD +BDA + DAB = 180°

90° + 30° + DAB = 180°

120° + DAB = 180°

⇒ ∠DAB = 180° 120° = 60°

Thus, the angles of ΔABD are 30°, 60° and 90°.

Therefore, the ratio of the lengths of the sides of ΔABD is 1::2.

Area of the rectangle = Length × Breadth

= AB × BD

#### Question 1:

Without actually drawing figures or looking up tables, can you arrange the numbers below is ascending order?

sin 1°, cos 1°, sin 2°, cos 2°

The value of sine increases as the measure of the angle increases.

sin 1° < sin 2°

The value of cosine decreases as the measure of the angle decreases.

cos 2° < cos 1°

From 0° to 45°, the value of cosine is greater than the value of sine and after that the value of sine is greater than the value of cosine.

sin 2° < cos 2°

sin 1° < sin 2° < cos 2° < cos 1°

Thus, the given numbers can be arranged in ascending order as follows:

sin 1°, sin 2°, cos 2°, cos 1°

#### Question 2:

The lengths of two sides of a triangle are 6 centimetres and 4 centimetres; and the angle between them is 130°. What is its area?

Let the given triangle be ABC with AB = 4 cm, BC = 6 cm and ABC = 130°.

Construction: Produce CB and draw AD perpendicular to CB produced.

By construction, we have ADB = 90°

We know that the sum of angles of a linear pair is 180°.

∴ ∠ABC + ABD = 180°

130° + ABD = 180°

⇒ ∠ABD = 180° 130° = 50°

AD = AB × Sin 50° … (1)

We know that Sin 50° 0.7660 (From table)

Putting the value of Sin 50° in (1):

= 3.064 cm

Area of a triangle =

Therefore, the area of ΔABC is approximately equal to 9.912 cm2.

#### Question 3:

One angle of a triangle is 110° and the side opposite to it is 4 centimetres long. What is its circumradius?

Let the given triangle be ABC with ABC = 110° and AC = 4 cm.

We need to find the circumradius for this triangle, so we make ΔABC in such a way that points A, B and C lie on the circle.

Construction: Mark a point D such that line segment CD passes through the centre O of the circle.

Join points A and D.

As line segment CD passes through the centre of the circle, CD is the diametre.

We know that angle in a semicircle is a right angle.

∴ ∠DAC = 90°

Now, ABCD is a cyclic quadrilateral and the sum of opposite angles of a cyclic quadrilateral is 180°.

∴ ∠ADC + ABC = 180°

⇒ ∠ADC + 110° = 180°

⇒ ∠ADC = 180° 110° = 70°

In ΔDAC:

(1)

We know that Sin 70° 0.9397 (From table)

Putting the value in (1):

#### Question 4:

Two sides of a triangle are 7 and 6 centimetres long and the angle between them is 140°. What is the length of the third side?

Let the given triangle be ABC with AB = 6 cm, BC = 7 cm and ABC = 140°.

Construction: Produce CB and draw AD perpendicular to CB.

By construction, we have ADB = 90°

We know that the sum of angles of a linear pair is 180°.

∴∠ABC + ABD = 180°

140° + ABD = 180°

⇒ ∠ABD = 180° 140° = 40°

In ΔABC:

AD = AB × Sin 40° … (1)

We know that Sin 40° 0.6428 (From table)

Putting the value in (1):

= 3.8568 cm

DB = AB × Cos 40° … (2)

We know that Cos 40° 0.7660 (from table)

Putting the value in equation (2):

DB 6 × 0.7660 cm

= 4.596 cm

DC = DB + BC

(4.596 + 7) cm

= 11.596 cm

Thus, the length of the third side of the triangle is approximately equal to12.22 cm.

#### Question 5:

Two sides of a parallelogram are of length 6 centimetres and 4 centimetres and the angle between them is 35°. What are the lengths of its diagonals?

Let the given parallelogram be ABCD with DC = 6 cm, AD = 4 cm and ADC = 35°.

Construction: Draw AE perpendicular to DC and join AC.

In ΔAED:

AE = AD × Sin 35° … (1)

We know that Sin 35° 0.5736 (From table)

Putting the value in (1):

AE 4 × 0.5736 cm

= 2.2944 cm

DE = AD × cos 35° … (2)

We know that cos 35° 0.8192 (From table)

Putting the value in (2):

DE 4 × 0.8192 cm

= 3.2768 cm

EC = DC DE

(6 3.2768) cm

= 2.7232 cm

Now, in ΔAEC:

Now, consider the parallelogram again.

Construction: Produce DC and draw BF perpendicular to DC. Join BD.

Also, AE = BF = 2.2944 cm

In ΔBFC:

CF = BC × cos 35° … (2)

We know that cos 35° 0.8192 (From table)

Putting the value in (2):

CF 4 × 0.8192 cm

= 3.2768 cm

DF = DC + CF

(6 + 3.2768) cm

= 9.2768 cm

Now, in ΔBFD:

Thus, the length of the diagonals of the given parallelogram is 3.56 cm and 9.56 cm.

#### Question 1:

How many rhombuses can we draw with one diagonal 5 centimetres long and one angle 50°? What are their areas?

We can draw two rhombuses with one diagonal 5 cm long and one angle 50°, explained in the following cases.

Case I:

Let ABCD be a rhombus with AC = 5 cm, ADC = 50°.

Let the diagonals AC and BD intersect each other at point O.

We know that diagonals of a rhombus are perpendicular bisectors of each other.

Also, diagonals of a rhombus bisect the vertex angles.

∴ ∠DOC = 90°

OC =

CDO =

In ΔDOC:

DO = … (1)

We know that tan 25° 0.4663 (From table)

Putting the value in (1):

DO cm

BD = 2 × DO

2 × 5.36 cm

= 10.72 cm

Thus, with the length of both the diagonals and measure of one angle, we can draw a unique rhombus.

Area of the rhombus =

Case 2:

Let ABCD be a rhombus with BD = 5 cm, ADC = 50°.

Let the diagonals AC and BD intersect each other at point O.

We know that diagonals of a rhombus are perpendicular bisectors of each other.

Also, diagonals of a rhombus bisect the vertex angles.

∴ ∠DOC = 90°

OD =

CDO =

In ΔDOC:

OC = DO × tan 25° … (1)

We know that tan 25° 0.4663 (From table)

Putting the value in (1):

OC 2.5 × 0.4663 cm

= 1.17 cm

AC = 2 × OC

2 × 1.17 cm

= 2.34 cm

Thus, with the length of both the diagonals and measure of one angle, we can draw a unique rhombus.

Area of the rhombus =

#### Question 2:

A ladder leans against a wall with its foot 2 metres away from the wall and it makes a 40° angle with the ground. How high is the top of the ladder from the ground?

Let AC be the ladder which leans against the wall AB.

BC is the distance between the foot of the ladder and the wall.

So, in ΔABC, BC = 2 cm, ABC = 90° and ACB = 40°.

In ΔABC:

AB = BC × tan 40° … (1)

We know that tan 40° 0.8391 (From table)

Putting the value in (1):

AB 2 × 0.8391cm

= 1.6782 cm

Thus, the top of the ladder is at a distance of approximately 1.6782 cm from the ground.

#### Question 4:

The vertical lines in the figure below are drawn 1 centimetres apart.

Prove that their heights are in arithmetic sequence.

What is the common difference?

Given: BC = CD = DE = 1 cm

Let AB = x cm

AC = (x + 1) cm, AD = (x + 2) cm, AE = (x + 3) cm

In ΔABF, ABF = 90°

In ΔACG, ACG = 90°

In ΔAEI, AEI = 90°

The sequence of the heights is , + 0.8391, + 2(0.8391), + 2(0.8391), …

Here, first term =

Second term = + 0.8391

Third term = + 2(0.8391)

Second term First term = + 0.8391 = 0.8391

Third term Second term = + 2(0.8391) 0.8391 = 0.8391

Since, third term second term = second term first term, the sequence of heights is an arithmetic sequence with common difference 0.8391.

#### Question 3:

Three rectangles are cut along their diagonals and the triangles so got are rearranged to form a regular pentagon as shown below:

Find the dimensions of the rectangles.

Given: ABCDE is a regular pentagon made by using some pieces of rectangles cut along its diagonal.

We know that all the sides of a regular pentagon are equal.

AB = BC = CD = DE = AE = 30 cm

Measure of each angle of a pentagon = 108°

∴ ∠AED = 108°

Measure of each angle of a rectangle = 90°

⇒ ∠AHE = DHE = 90°

In ΔAED

AE = DE

∴ ∠EDA = EAD (Angles opposite to equal sides are equal in measure.)

Using angle sum property in ΔAED:

AED + EDA + DAE = 180°

108° + 2EDA = 180°

2EDA = 180° 108°

2EDA = 72°

⇒ ∠EDA = 36°

In ΔEHD:

Therefore, the dimensions of the smaller rectangle are 24.27 cm and 17.63 cm.

= 108° 36°

= 72°

CF = FD = =15 cm

In ΔAFD:

Therefore, the dimensions of the larger rectangle are 46.165 cm and 15 cm.

#### Question 1:

The length of the shadow of a tree is 18 metres, when the sun is at an elevation of 40°. What is the height of the tree?

Let AB be the height of the tree and BC be the length of the shadow.

Given: BC = 18 m and ACB = 40°

In ΔABC, ABC = 90°

AB = BC × tan 40°

AB 18 × 0.8391 cm

= 15.1038 m

Therefore, the height of the tree is approximately 15.10 m.

#### Question 2:

A man 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°. On climbing to the top of the tower, he sees the top of the hill at an elevation of 50°. Compute the heights of the hill and the tower.

Let EG be the height of the tower and EF be the height of the man standing at the foot of the tower.

Let AD be the height of the hill.

Let AFC be the angle of elevation made by the man’s eye while seeing the hill from the foot of the tower.

Let GH be the position of the man when he reaches the top of the tower to see the hill.

Let AHB be the angle made by the man’s eye while seeing the hill from the top of the tower.

Let DE be the distance between the tower and the hill.

Let AB = y m, GF = x m

The figure for the given situation with dimensions can be made as follows:

In ΔABH:

Now in ΔACF:

Height of the hill = (y + x + 1.75 + 1.75) m

(47.672 + 19.862 + 1.75 + 1.75) m

= 71.034 m

Height of the tower = (x + 1.75) m

(19.862 + 1.75) m

= 21.612 m

Thus, the height of the hill and the tower is approximately equal to 71.034 m and 21.612 m.

#### Question 3:

A boy 1.5 metres tall, sees the top of a building under construction at an elevation of 30°. The building is completed, adding 10 more metres to its height; and then the boy sees the top at an elevation of 60° from the same spot. What is the total height of the completed building?

Let AB be the height of the child.

Let EC be the height of under constructed building and FC be the height of the complete building.

Let EAD be the angle of elevation made by the boy’s eye while seeing the under constructed building.

Let FAD be the angle of elevation made by the boy’s eye while seeing the completed building.

Let BC be the distance between the building and the child.

Let ED = x m and BC = y m

The figure for the given situation with dimensions can be made as follows:

In ΔAED:

Now in ΔAFD:

From equation (1) and equation (2), we get:

Total height of the completed building = (10 + x + 1.5) m

(10 + 4.96 + 1.5) m

= 16.46 m

Thus, the total height of the completed building is approximately equal to 16.46 m.

#### Question 4:

A man 1.8 metres tall, looking down from the top of a telephone tower sees the top of a building 10 metres high at an angle of depression 40° and the foot of the building at an angle of depression 60°. What is the height of the tower? How far is it away from the building?

Let AF be the height of the man.

Let FC be the height of the telephone tower on which the man is standing.

Let ED be the height of the building.

Let CD be the distance between the telephone tower and the building.

Let XAE be the angle of depression made by the man’s eye while seeing the top of the building.

Let XAD be the angle of depression made by the man’s eye while seeing the top of the building.

The figure for the given situation with dimensions can be made as follows:

In ΔAEB:

Now, in ΔACD:

From equation (1) and equation (2), we get:

Height of the telephone tower = (10 + x) m

= (10 + 7.6) m

= 17.6 m

Putting the value of x in equation (1), we get:

Thus, the height of the telephone tower and the distance of the telephone tower from the building are 17.6 m and 11.2 m respectively.

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