Mathematics Part I Solutions Solutions for Class 10 Math Chapter 4 Trigonometry are provided here with simple step-by-step explanations. These solutions for Trigonometry are extremely popular among Class 10 students for Math 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 Math 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 Math are prepared by experts and are 100% accurate.

#### Page No 77:

#### 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?

#### Answer:

Let *ABCD* be the given parallelogram.

*DC* = 6 cm and ∠*ADC* = 60°

Area of parallelogram *ABCD* = 30 cm^{2}

**Construction:** Draw *AE *perpendicular to *DC*.

Area of a parallelogram = Base × Height

⇒ 30 cm^{2} = 6 cm × *AE*

⇒ *AE* =

Using angle sum property in Δ*AED*:

∠*ADE* + ∠*DEA* + ∠*EAD* = 180°

⇒ 60° + 90° + ∠*EAD* = 180°

⇒ 150° + ∠*EAD* = 180°

⇒ ∠*EAD* = 180° − 150° = 30°

Thus, the angles of Δ*AED *are 30°, 60° and 90°.

Therefore, the ratio of the lengths of the sides of Δ*AED* 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.

#### Page No 77:

#### Question 2:

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

#### Answer:

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, Δ*ABC *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 Δ*ABC*.

*AD* is the perpendicular bisector of side *BC*.

⇒ ∠*ADB* = 90°

Using angle sum property in Δ*ADB*:

∠*ABD* + ∠*BDA* + ∠*DAB* = 180°

⇒ 60° + 90° + ∠*DAB* = 180°

⇒ 150° + ∠*DAB* = 180°

⇒ ∠*DAB* = 180° − 150° = 30°

Thus, the angles of Δ*ADB* are 30°, 60° and 90°.

Therefore, the ratio of the lengths of the sides of Δ*ADB* 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 .

#### Page No 77:

#### Question 3:

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

#### Answer:

Let Δ*ABC* be the given right-angled triangle, right-angled at *B*.

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

Using angle sum property in Δ*ABC*:

∠*ABC* + ∠*BCA* + ∠*CAB* = 180°

⇒ 90° + 30° + ∠*CAB* = 180°

⇒ 120° + ∠*CAB* = 180°

⇒ ∠*CAB* = 180° − 120° = 60°

Thus, the angles of Δ*ABC* are 30°, 60° and 90°.

Therefore, the ratio of the lengths of the sides of Δ*ABC* is 1::2.

#### Page No 78:

#### Question 1:

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

What is the diameter of the circle?

#### Answer:

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 Δ*ABC**:*

∠*ABC* + ∠*BCA* + *∠**CAB** *= 180°

⇒ 90° + ∠*BCA* + 30° = 180°

⇒ 120° + ∠*BCA* = 180°

⇒ ∠*BCA* = 180° − 120° = 60°

Thus, the angles of Δ*ABC* are 30°, 60° and 90°.

Therefore, the ratio of the lengths of the sides of Δ*ABC* is 1::2.

_{AC = 2 × BC = 2 × 2.5 cm = 5 cm}

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

#### Page No 78:

#### Question 2:

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

#### Answer:

**Construction: **Draw *OD* perpendicular to *AB*.

In Δ*AOB*,* OA *= *OB*

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

Using angle sum property in Δ*AOB*:

∠*AOB* + ∠*OBA* + ∠*BAO** *= 180°

⇒ 120° + 2∠*BAO* = 180°

⇒ 2∠*BAO* = 180° − 120°

⇒ 2∠*BAO* = 60°

⇒ ∠*BAO* = 30°

Now, using angle sum property in Δ*ADO*:

∠*AOD** *+ ∠*ODA* + ∠*DAO* = 180°

⇒ ∠*AOD* + 90° + 30° = 180°

⇒ ∠*AOD* + 120° = 180°

⇒ ∠*AOD* = 180° − 120° = 60°

Thus, the angles of Δ*ADO* are 30°, 60° and 90°.

Therefore, the ratio of the lengths of the sides of Δ*ADO* is 1::2.

⇒ *AD* =

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

⇒ *AB *= 2*AD* =

Therefore, the length of the chord is

#### Page No 78:

#### 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?

#### Answer:

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 Δ*EFD*:

∠*EFD* + ∠*FDE* + ∠*DEF** *= 180°

⇒ 90° + 30° + ∠*DEF** *= 180°

⇒ 120° + ∠*DEF* = 180°

⇒ ∠*DEF* = 180° − 120° = 60°

Thus, the angles of Δ*EFD* are 30°, 60° and 90°.

Therefore, the ratio of the lengths of the sides of Δ*EFD* 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

#### Page No 79:

#### Question 1:

Compute the lengths of all sides of the quadrilateral below:

#### Answer:

Using angle sum property in Δ*ABC*:

∠*ABC* +∠*BCA* + ∠*CAB* = 180°

⇒ 90° + 30° + ∠*CAB** *= 180°

⇒ 120° + ∠*CAB** *= 180°

⇒ ∠*CAB* = 180° − 120° = 60°

Thus, the angles of Δ*ABC* are 30°, 60° and 90°.

Therefore, the ratio of the lengths of the sides of Δ*ABC* is 1::2.

Now, using angle sum property in Δ*ADC*:

∠*ADC* +∠*DCA* + ∠*CAD* = 180°

⇒ 90° + ∠*DCA** *+ 45° = 180°

⇒ 135° + ∠*DCA* = 180°

⇒ ∠*DCA** *= 180° − 135° = 45°

Thus, the angles of Δ*ADC* are 45°, 45° and 90°.

Therefore, the ratio of the lengths of the sides of Δ*ADC* is 1:1:

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

#### Page No 79:

#### Question 2:

Compute the area of the rectangle below:

#### Answer:

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 Δ*ABC*, ∠*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 Δ*ABC* are 30°, 60° and 90°.

Therefore, the ratio of the lengths of the sides of Δ*ABC* is 1::2.

Now, using angle sum property in Δ*ABD*:

∠*ABD** *+∠*BDA** *+ ∠*DAB** *= 180°

⇒ 90° + 30° + ∠*DAB* = 180°

⇒ 120° + ∠*DAB* = 180°

⇒ ∠*DAB** *= 180° − 120° = 60°

Thus, the angles of Δ*ABD* are 30°, 60° and 90°.

Therefore, the ratio of the lengths of the sides of Δ*ABD** *is 1::2.

Area of the rectangle = Length × Breadth

= *AB *× *BD *

#### Page No 86:

#### 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°

#### Answer:

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°

#### Page No 86:

#### 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?

#### Answer:

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°

In Δ*ADB*:

⇒ *AD *= *AB* × Sin 50° … (1)

We know that Sin 50° ≈ 0.7660 (From table)

Putting the value of Sin 50° in (1):

*AD* ≈ (4 × 0.7660) cm

= 3.064 cm

Area of a triangle =

Therefore, the area of Δ*ABC* is approximately equal to 9.912 cm^{2}.

#### Page No 86:

#### Question 3:

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

#### Answer:

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 Δ*ABC* 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 Δ*DAC*:

⇒ … (1)

We know that Sin 70° ≈ 0.9397 (From table)

Putting the value in (1):

#### Page No 86:

#### 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?

#### Answer:

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 Δ*ABC*:

⇒ *AD* = *AB* × Sin 40° … (1)

We know that Sin 40° ≈ 0.6428 (From table)

Putting the value in (1):

*AD* ≈ 6 × 0.6428

= 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

Now, in Δ*ADC*:

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

#### Page No 86:

#### 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?

#### Answer:

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 Δ*AED*:

⇒ *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 Δ*AEC*:

Now, consider the parallelogram again.

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

As *AD* is parallel to *BC*, ∠*ADC* = ∠*BCF* = 35°.

Also, *AE* = *BF* = 2.2944 cm

In Δ*BFC*:

⇒ *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 Δ*BFD*:

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

#### Page No 89:

#### Question 1:

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

#### Answer:

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 Δ*DOC*:

⇒ *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 Δ*DOC*:

⇒ *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 =

#### Page No 89:

#### 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?

#### Answer:

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 Δ*ABC*,* BC *= 2 cm, ∠*ABC* = 90° and ∠*ACB** *= 40°.

In Δ*ABC*:

⇒ *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.

#### Page No 89:

#### 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?

#### Answer:

Given:** ***BC* = *CD* =* DE* = 1 cm

Let *AB* = *x *cm

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

In Δ*ABF*, ∠*ABF* = 90°

In Δ*ACG*, ∠*ACG* = 90°

In Δ*ADH*, ∠*ADH** *= 90°

In Δ*AEI*, ∠*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.

#### Page No 89:

#### 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.

#### Answer:

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 Δ*AED*:

*AE* =* DE*

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

Using angle sum property in Δ*AED*:

∠*AED* + ∠*EDA* + ∠*DAE** *= 180°

⇒ 108° + 2∠*EDA* = 180°

⇒ 2∠*EDA* = 180° − 108°

⇒ 2∠*EDA* = 72°

⇒ ∠*EDA* = 36°

In Δ*EHD*:

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

Now, ∠*ADC *= ∠*EDC** *− ∠*EDH** *

= 108° − 36°

= 72°

*CF* = *FD* = =15 cm

In Δ*AFD*:

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

#### Page No 93:

#### 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?

#### Answer:

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

Given: *BC* = 18 m and ∠*ACB** *= 40°

In Δ*ABC*, ∠*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.

#### Page No 93:

#### 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.

#### Answer:

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 Δ*ABH*:

Now in Δ*ACF*:

∴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.

#### Page No 93:

#### 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?

#### Answer:

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 Δ*AED*:

Now in Δ*AFD*:

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.

#### Page No 93:

#### 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?

#### Answer:

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 Δ*AEB*:

Now, in Δ*ACD*:

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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