Circles Chapter 9 : NCERT Exemplar for Class 10

NCERT Exemplar Solutions for Class 10 Maths Chapter 9 Circles are provided here for students to download in PDF and prepare well for the CBSE Board exams.Board exams are something called a tough bridge between students and examinations which can be ace by a proper planning and solution structure of every chapter, These exemplar problems and solutions are designed by our expert faculty in accordance with the latest CBSE Syllabus (2024-2025). Using these solved questions with detailed answers to all the chapter questions, students will be able to clear all their doubts regarding the topics and study effectively for the exams.

They can also refer to these study materials to prepare for competitive exams in the future. Simply Acad provides a page for every student by providing them a set of solutions in a form of pdf that is free. In Chapter 9, students will explore the topic of “Circle” and learn about the tangent and its existence in a circle. They will also come across several examples and activities in the chapter, which will help them understand and solve different types of questions. Thus, to facilitate easy learning and help students grasp all the concepts of circles properly,

See below,practice hard and achieve the milestone of your good grades in boards!!

Question 1. when cos A = 4/5, the value

for tan A will

(A) 5/3

(B) ¾

(C) 4/3

(D) 3/5

Answer 1. (B) 3/4

Explanation: According to the question,

cos A = 4/5 …(1)

We know,

tan A = sinA/cosA

To find the value of sin A,

We have the equation,

sin2 θ +cos2 θ =1

So, sin θ = √ (1-cos2 θ)

Then,

sin A = √ (1-cos2 A) …(2)

sin2 A = 1-cos2 A

sin A = √(1-cos2 A)

Substituting equation (1) in (2),

We get,

Sin A = √(1-(4/5)2)

= √(1-(16/25))

= √(9/25)

= ¾

Therefore,

tan A = (3/5) x (5/4) = 3/4

 

 

Question 2. When sin A = (1/2), the value of cot A will be

(A) √3

(B) 1

(C) √3/2

(D) 1/√3

Answer 2. (A) √3

Explanation: According to the question,

Sin A = ½ … (1)

We know that,

cot A = 1/tanA = cosA/sinA … (2)

To find the value of cos A.

We have the equation,

sin2 θ +cos2 θ =1

So, cos θ = √(1-sin2 θ)

Then,

cos A = √(1-sin2 A) … (3)

cos2 A = 1-sin2 A

cos A = √ (1-sin2 A)

Substituting equation 1 in 3, we get,

cos A = √(1-1/4) = √(3/4) = √3/2

Substituting value of sin A as well as cos A in equation 2, we get

cot A = (√3/2) × 2 = √3

 

Question 3. The value for the expression [cosec (75° + θ) – sec (15° – θ) – tan (55° + θ) + cot (35° – θ)] will be_______.

(A) 1

(B) 0

(C) – 1

(D) 3 / 2

 

Answer 3. (B) 0

Explanation: As per question,

We have to find the value for the equations,

cosec(75°+θ) – sec(15°-θ) – tan(55°+θ) + cot(35°-θ)

= cosec[90°-(15°-θ)] – sec(15°-θ) – tan(55°+θ) + cot[90°-(55°+θ)]

Since, cosec (90°- θ) = sec θ

thus, cot(90°-θ) = tan θ

We observe,

= sec(15°-θ) – sec(15°-θ) – tan(55°+θ) + tan(55°+θ)

= 0

 

Question 4. if sinθ = a b, then cosθ will equal to_______.

(A) a/√(b2– a2)

(B) b/a

(C) √(b2-a2)/b

(D) b/√(b2-a2)

 

Answer 4. (C) √(b2 – a2)/b

Explanation: According to the question,

sin θ =a/b

We know that, sin2 θ +cos2 θ =1

sin2 A = 1-cos2 A

sin A = √(1-cos2 A

thus, cos θ = √(1-a2/b2 ) = √((b2-a2)/b2 ) = √(b2-a2 )/b

Hence, cos θ = √(b2 – a2 )/b

 

 

Question 5. When cos (α + β) = 0, sin (α – β) can be reduced to______.

(A)sin 2α

(B) cos 2β

(C) sin α

(D) cos β

 

Answer 5. (B) cos 2β

Explanation: According to the question,

cos(α+β) = 0

Since, cos 90° = 0

We can write,

cos(α+β)= cos 90°

By comparing the cosine equation on L.H.S and R.H.S,

We observe,

(α+β)= 90°

α = 90°-β

then, we need to reduce sin (α -β ),

So, we take,

sin(α-β) = sin(90°-β-β) = sin(90°-2β)

sin(90°-θ) = cos θ

thus, sin(90°-2β) = cos 2β

Therefore, sin(α-β) = cos 2β

 

Question 6. The value of (tan1° tan2° tan3° … tan89°) will be________.

(A) 0

(B) 1

(C) 2

(D) ½

 

Answer 6. (B) 1

Explanation: tan 1°. tan 2°.tan 3° …… tan 89°

= tan1°.tan 2°.tan 3°…tan 43°.tan 44°.tan 45°.tan 46°.tan 47°…tan 87°.tan 88°.tan 89°

hence, tan 45° = 1,

= tan1°.tan 2°.tan 3°…tan 43°.tan 44°.1.tan 46°.tan 47°…tan 87°.tan 88°.tan 89°

= tan1°.tan 2°.tan 3°…tan 43°.tan 44°.1.tan(90°-44°).tan(90°-43°)…tan(90°-3°). tan(90°-2°).tan(90°-1°)

so, tan(90°-θ) = cot θ,

= tan1°.tan 2°.tan 3°…tan 43°.tan 44°.1.cot 44°.cot 43°…cot 3°.cot 2°.cot 1°

thus, tan θ = (1/cot θ)

= tan1°.tan 2°.tan 3°…tan 43°.tan 44°.1. (1/tan 44o). (1/tan 43o)… (1/tan 3o). (1/tan 2o). (1/tan 1o)

= (tan1° x 1/tan1°).(tan2° x 1/tan2°)…(tan44° x 1/tan44°)

= 1

thus, tan 1°.tan 2°.tan 3° …… tan 89° = 1

 

Question 7. If cos 9α = sinα and 9α < 90°, then the value of tan5α is________.

(A) 1/√3

(B) √3

(C) 1

(D) 0

 

Answer 7. (C) 1

Explanation: According to the question,

cos 9∝ = sin ∝ and 9∝<90°

i.e. 9α is an acute angle

We know that,

sin(90°-θ) = cos θ

So,

cos 9∝ = sin (90°-∝)

Since, cos 9∝ = sin(90°-9∝) and sin(90°-∝) = sin∝

Thus, sin (90°-9∝) = sin∝

90°-9∝ =∝

10∝ = 90°

∝ = 9°

Substituting ∝ = 9° in tan 5∝, we get,

tan 5∝ = tan (5×9) = tan 45° = 1

∴, tan 5∝ = 1

 

Question 8. Mention true or false: tan 47o/cot 43° = 1

Answer 8.

True

Justification:-

Since, tan (90° -θ) = cot θ

tan 47o/cot 43° = tan (90o– 43o)/cot 43°

tan 47o/cot 43° = cot 43°/cot 43° =1

tan 47o/cot 43° = 1

 

Question 9. Is this statement True or False? The value for the expression (cos223° – sin267°) is positive.

Answer 9.

False

clarification:

Since, (a2-b2) = (a+b)(a-b)

cos2 23° – sin2 67° =(cos 23°+sin 67°)(cos 23°-sin 67°)

= [cos 23°+sin(90°-23°)] [cos 23°-sin(90°-23°)]

= (cos 23°+cos 23°)(cos 23°-cos 23°) (∵sin(90°-θ) = cos θ)

= (cos 23°+cos 23°).0

= 0, that is neither positive nor negative

 

Question 10. Is this statement True or False? The value for the expression (sin 80° – cos 80°) is negative.

Answer 10.

False

Justification:

We know that

sin θ increases when 0° ≤ θ ≤ 90°

cos θ decreases when 0° ≤ θ ≤ 90°

And (sin 80°-cos 80°) = (increasing value-decreasing value)

= a positive value.

Therefore, (sin 80°-cos 80°) > 0.

 

Question 11. Is this statement True or False? If cosA + cos2A = 1, then sin2A + sin4A = 1.

Answer 11.

True

Justification:

According to the question,

cos A+cos2 A = 1

i.e., cos A = 1- cos2 A

Since,

sin2 θ+cos2 θ = 1

sin2 θ = 1- cos2 θ)

We get,

cos A = sin2 A …(1)

Squaring L.H.S and R.H.S,

cos2 A = sin4 A …(2)

To find sin2A+sin4 A=1

Adding equations (1) and (2),

We get

sin2A + sin4 A= cos A + cos2 A

Therefore, sin2A+ sin4 A = 1

 

Question 12. Is this statement True or False? (tan θ + 2) (2 tan θ + 1) = 5 tan θ + sec2 θ.

Answer 12.

False

Justification:

L.H.S = (tan θ+2) (2 tan θ+1)

= 2 tan2 θ + tan θ + 4 tan θ + 2

= 2 tan2θ+5 tan θ+2

Since, sec2 θ – tan2 θ = 1, we get, tan2θ = sec2 θ-1

= 2(sec2 θ-1) +5 tan θ+2

= 2 sec2 θ-2+5 tan θ+2

= 5 tan θ+ 2 sec2 θ ≠R.H.S

∴, L.H.S ≠ R.H.S

 

Question 13. Prove this statement. If tan A = ¾, then sinA cosA = 12/25

Answer 13.

According to the question,

tan A = ¾

We know,

tan A = perpendicular/ base

So,

tan A = 3k/4k

Where,

Perpendicular = 3k

Base = 4k

Source: Internet

Using Pythagoras Theorem,

(hypotenuse)2 = (perpendicular)2 + (base)2

(hypotenuse)2 = (3k)2+ (4k)2 = 9k2+16k2 = 25k2

hypotenuse = 5k

To find sin A and cos A,

sinA = perpendicular/hypotenuse = 3k/5k = (3/5)

cosA = base/hypotenuse = 4k/5k = (4/5)

multiplying sinA and cos A,

sinA and cosA = (3/5)x(4/5) = 12/25

Hence, proved.

 

Question 14. Prove – (√3+1) (3 – cot 30°) = tan3 60° – 2 sin 60°

Answer 14.

L.H.S: (√3 + 1) (3 – cot 30°)

= (√3 + 1) (3 – √3) [∵cos 30° = √3]

= (√3 + 1) √3 (√3 – 1) [∵(3 – √3) = √3 (√3 – 1)]

= ((√3)2– 1) √3 [∵ (√3+1)(√3-1) = ((√3)2 – 1)]

= (3-1) √3

= 2√3

Similarly, solving R.H.S: tan3 60° – 2 sin 60°

Since, tan 60o = √3 and sin 60o = √3/2,

We get,

(√3)3 – 2.(√3/2) = 3√3 – √3

= 2√3

Therefore, L.H.S = R.H.S

Hence, proved.

 

Question 15. If 1 + sin2θ = 3sinθ cosθ , then prove that tanθ = 1 or ½.

Answer 15.

Given: 1+sin2 θ = 3 sin θ cos θ

Dividing L.H.S and R.H.S equations with sin2 θ,

We get,

= (1+sin2 θ)/sin2 θ = 3 sin θ cos θ/sin2 θ

= (1/sin2 θ) + 1 = 3cos θ/sin θ

cosec2 θ + 1 = 3 cot θ

Since,

cosec2 θ – cot2 θ = 1 ⇒ cosec2 θ = cot2 θ +1

⇒ cot2 θ +1+1 = 3 cot θ

⇒ cot2 θ +2 = 3 cot θ

⇒ cot2 θ –3 cot θ +2 = 0

Splitting the middle term and then solving the equation,

⇒ cot2 θ – cot θ –2 cot θ +2 = 0

⇒ cot θ(cot θ -1)–2(cot θ +1) = 0

⇒ (cot θ – 1)(cot θ – 2) = 0

⇒ cot θ = 1, 2

Since,

tan θ = 1/cot θ

tan θ = 1, ½

Hence, proved.

 

Question 16. Given that sinθ + 2cosθ = 1, then prove that 2sinθ – cosθ = 2.

Answer 16.

Given: sin θ +2 cos θ = 1

Squaring on both sides,

(sin θ +2 cos θ)2 = 1

⇒ sin2 θ + 4 cos2 θ + 4sin θcos θ = 1

Since, sin2 θ = 1 – cos2 θ and cos2 θ = 1 – sin2 θ

⇒ (1 – cos2 θ) + 4(1 – sin2 θ) + 4sin θcos θ = 1

⇒ 1 – cos2 θ + 4 – 4 sin2 θ + 4sin θcos θ = 1

⇒ – 4 sin2 θ – cos2 θ + 4sin θcos θ = – 4

⇒ 4 sin2 θ + cos2 θ – 4sin θcos θ = 4

We know that,

a2+ b2 – 2ab = ( a – b)2

So, we get,

(2sin θ – cos θ)2 = 4

⇒ 2sin θ – cos θ = 2

Hence proved.

 

Question 17. The circus artist is climbing a 20 m long rope that is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole when the angle made by the rope with the ground level is 30°. (see fig.)

 

Answer 17.

The length for the rope is 20 m, and the angle made by the rope with the ground level is 30°.

Given that: AC = 20 m and angle C = 30°

To Find out: Height of the pole

Let AB be the vertical pole

In the right ΔABC, using the sine formula

sin 30° = AB/AC

Using the value for sin 30 degrees is ½, we have

1/2 = AB/20

AB = 20/2

AB = 10

so the height of the pole is 10 m.

 

Question 18. A tree breaks due to a storm as well as the broken part bends so that the top of the tree touches the ground, making an angle of 30° with it. The distance between the foot of a tree and the point where the top touches the ground is 8 m. Find out the tree’s height.

Answer 18.

Using the given instructions, draw a figure. Let’s AC is the broken part of the tree. Angle C = 30°

BC = 8 m

To Find out: Height of the tree, which is AB

From the figure: The total height of the tree is the sum of AB and AC, i.e. AB+AC

In the right ΔABC,

Using Cosine and tangent angle,

cos 30° = BC/AC

We know that cos 30° = √3/2

√3/2 = 8/AC

AC = 16/√3 …(1)

Also,

tan 30° = AB/BC

1/√3 = AB/8

AB = 8/√3 ….(2)

thus, total height of the tree = AB + AC = 16/√3 + 8/√3 = 24/√3 = 8√3 m.

 

Question 19. A contractor plans to install the two slides for children to play in a park. For the children below the age of 5 years, she prefers to have a slide whose top is at the height of 1.5 m as well as is inclined at an angle of 30° to the ground, whereas, for elder children, she wants to have a steep slide at the height of 3m, and inclined at the angle of 60° to ground. What should be the length of the slide in each case?

Answer 19.

As per the contractor’s plan,

 

Ages	Height of sides	Inclined Angle
Below five year	1.5m	30 degree
Above five year	3m	60 degree

 

Let ABC is the slide inclined at 30° with the length AC, and PQR is the slide inclined at

60° with length PR.

To Find: AC and PR

In the right ΔABC,

sin 30° = AB/AC

1/2 = 1.5/AC

AC = 3

Also,

In the right ΔPQR,

sin 60° = PQ/PR

⇒ √3/2 = 3/PR

⇒ PR = 2√3

Hence, the length of the slide below 5 = 3 m and

the length of the slide for elderly children = 2√3 m

 

Question 20. The angle for the elevation of the top of a tower from the point on the ground 30 m away from the foot of the tower is 30°. Find out the height of the tower.

Answer 20.

Let AB be the tower’s height, and C is the point of elevation 30 m away from the foot of the tower.

 

To Find: AB (Height of the tower)

In right ABC

tan 30° = AB/BC

1/√3 = AB/30

⇒ AB = 10√3

Here, the height of the tower is 10√3 m.

 

Question 21. A kite is flying at the height of 60 m above ground. The string attached to the kite is temporarily tied to the point on the ground. The inclination of a string with the ground is 60°. Find out the string’s length, assuming there is no slack in the string.

Answer 21.

Draw a figure based on the given instruction,

Lets BC = height of the kite from the ground, BC = 60 m

AC = Inclined length of string from the ground and

A is the point where the string of the kite is tied.

To Find the Length of the string from the ground, i.e. the value of AC

From the above figure,

sin 60° = BC/AC

⇒ √3/2 = 60/AC

⇒ AC = 40√3 m

hence, the string length from the ground is 40√3 m.

 

Question 22. 1.5 m tall boy stood a few miles from the 30 m tall building. The elevation angle from his eyes to the top of a building increases from 30° to 60° as he walks toward the building. Find out the distance he walked towards the building.

Answer 22.

Lets the boy initially stand at point Y with an inclination of 30°, and then he approaches the building to point X with an inclination of 60°.

To Find out: The distance boy walked towards the building, i.e. XY

From the figure,

XY = CD.

Height of building = AZ = 30 m.

AB = AZ – BZ = 30 – 1.5 = 28.5

The measure of AB is 28.5 m

In the right ΔABD,

tan 30° = AB/BD

1/√3 = 28.5/BD

BD = 28.5√3 m

Again,

In the right ΔABC,

tan 60° = AB/BC

√3 = 28.5/BC

BC = 28.5/√3 = 28.5√3/3

hence, the length BC is 28.5√3/3 m.

XY = CD = BD – BC = (28.5√3-28.5√3/3) = 28.5√3(1-1/3) = 28.5√3 × 2/3 = 57/√3 = 19√3 m.

so, the distance the boy walked towards the building was 19√3 m.

 

Question 23. From a point on the ground, the angles of elevation of the bottom and the top of a

transmission towers fixed at top of a 20 m high building are 45° and 60°, respectively. Find the height of the tower.

Answer 23.

Let’s BC is the 20 m high building.

D is a point on the ground from where the elevation is taken.

Height of the transmission tower = AB = AC – BC

To Find: AB, Height of the tower

From the figure, In the right ΔBCD,

tan 45° = BC/CD

1 = 20/CD

CD = 20

Again,

In the right ΔACD,

tan 60° = AC/CD

√3 = AC/20

AC = 20√3

then, AB = AC – BC = (20√3-20) = 20(√3-1)

Height of the transmission tower = 20(√3 – 1) m.

 

Question 24. An 1.6 m tall statue stands on the top of the pedestal. From a point on the ground, the angle of elevation of the statue is 60°, and from the same point, the angle of elevation of the top of the pedestal is 45°. Find out the height of the pedestal.

Answer 24.

Let AB be the height of the statue.

D is the point on ground from here; the elevation is taken.

To Find: Height of pedestal = BC = AC-AB

From figure,

In the right triangle BCD,

tan 45° = BC/CD

1 = BC/CD

BC = CD …..(1)

 

Again,

In the right ΔACD,

tan 60° = AC/CD

√3 = ( AB+BC)/CD

√3CD = 1.6 + BC

√3BC = 1.6 + BC (use equation (1)

√3BC – BC = 1.6

BC(t√3-1) = 1.6

BC = [(1.6)(√3+1)]/[(√3-1)(√3+1)]

BC = [1.6(√3+1)]/(2) m

BC = 0.8(√3+1)

here, the height of the pedestal is 0.8(√3+1) m.

 

Question 25. The angle for elevation of the top of a building from the foot of the tower is 30°, as well as the angle of elevation of the top of the tower from the foot for the building, is 60°. When the tower is 50 m high, find out the height of the building.

Answer 25.

Let CD be the height of the tower. AB be the height of the building. BC is the distance in between the foot of the building and the tower. The elevation is 30 and 60 degrees from the tower and building.

In the right ΔBCD,

tan 60° = CD/BC

√3 = 50/BC

BC = 50/√3 …(1)

Again,

In the right ΔABC,

tan 30° = AB/BC

⇒ 1/√3 = AB/BC

Use the result obtained in the equation (1)

AB = 50/3

so the height of the building is 50/3 m.

 

Question 26. Two poles of equal height stand opposite each other on either side of the road, 80 m wide. From the point between them on the road, the elevation angles of the tops of the poles are 60° and 30°, respectively. Find out the height of the poles and the distance of the point from the poles.

Answer 26.

Let AB and CD be the poles of equal height.

O is the point between them from where the height elevation height is taken. BD is the distance in between the poles.

As per the above figure, AB = CD,

OB + OD = 80 m

then,

In the right ΔCDO,

tan 30° = CD/OD

1/√3 = CD/OD

CD = OD/√3 … (1)

Again,

In the right ΔABO,

tan 60° = AB/OB

√3 = AB/(80-OD)

AB = √3(80-OD)

AB = CD (Given)

√3(80-OD) = OD/√3 (Using equation (1))

3(80-OD) = OD

240 – 3 OD = OD

4 OD = 240

OD = 60

Putting value of OD in equation (1)

CD = OD/√3

CD = 60/√3

CD = 20√3 m

Also,

OB + OD = 80 m

⇒ OB = (80-60) m = 20 m

Thus, the height of the poles is 20√3 m, and distance from the point of elevation is 20 m and

60 m respectively.

 

Question 27. A TV tower stands vertically on a bank of a canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is 60°. From another point 20 m away from this point on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is 30° (see Fig. 9.12). Find the height of the tower and the width of the canal.

 

Answer 27. Given that AB is the height of the tower.

DC = 20 m (given)

As per the given diagram, In the right ΔABD,

tan 30° = AB/BD

1/√3 = AB/(20+BC)

AB = (20+BC)/√3 … (i)

Again,

In the right ΔABC,

tan 60° = AB/BC

√3 = AB/BC

AB = √3 BC … (ii)

From equations (i) and (ii)

√3 BC = (20+BC)/√3

3 BC = 20 + BC

2 BC = 20

BC = 10

Putting value of BC in equation (ii)

AB = 10√3

Tha implies that the tower’s height is 10√3 m and the width of the canal is 10 m.

 

Question 28. From the top of a 7 m high building, the angle of the elevation of the top of a cable tower is 60°, as well as the angle of depression of its foot, is 45°. Determine the height of the tower.

Answer 28.

Let AB be the building’s height of 7 m, and EC be the tower’s height.

A is the point from where the elevation of the tower is 60°, and the angle of depression of its foot is 45°.

EC = DE + CD

so, CD = AB = 7 m. and BC = AD

To Find: EC = height of tower

Design a figure based on the given instructions:

 

In the right ΔABC,

tan 45° = AB/BC

1= 7/BC

BC = 7

Since BC = AD

So AD = 7

Again, from the right triangle ADE,

tan 60° = DE/AD

√3 = DE/7

⇒ DE = 7√3 m

Now: EC = DE + CD

= (7√3 + 7) = 7(√3+1)

thus, the height of the tower is 7(√3+1) m.

 

Question 29. As observed from the top of the 75 m high lighthouse from sea level, the angles of depression of the two ships are 30° as well as 45°. When one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.

Answer 29.

Let AB be the lighthouse of a Height of 75 m. Let C, as well as D, be the positions of the ships.

30° and 45° are angles of depression from the lighthouse.

Draw the figure based on the given instructions:

To Find: CD = distance between two ships

Step 1: From the right triangle ABC,

tan 45° = AB/BC

1= 75/BC

BC = 75 m

Step 2: Form right triangle ABD,

tan 30° = AB/BD

1/√3 = 75/BD

BD = 75√3

Step 3: To find the measure of CD, use the results obtained in steps 1 and 2.

CD = BD – BC = (75√3 – 75) = 75(√3-1)

The distance between the two ships is 75(√3-1) m.

 

Question 30. A 1.2 m tall girl spots the balloon moving with the wind in a horizontal line at 88.2 m from ground height. The angle of elevation for the balloon from the eyes of the girl at any instant is 60°. After some time, the angle of elevation reduces to 30° (see Fig.). Find the distance travelled by balloon during the interval.

 

Answer 30.

Let the initial position for the balloon be A and the final position be B.

Height of the balloon above the girl height = 88.2 m – 1.2 m = 87 m.

To Find out: Distance travelled by the balloon = DE = CE – CD

Let us redesign given figure as per our convenient

 

1-Step: In the right ΔBEC,

tan 30° = BE/CE

1/√3= 87/CE

CE = 87√3

2-Step:

In the right ΔADC,

tan 60° = AD/CD

√3= 87/CD

CD = 87/√3 = 29√3

3-Step:

DE = CE – CD = (87√3 – 29√3) = 29√3(3 – 1) = 58√3

Distance travel by balloon = 58√3 m.

 

Question 31. A straight highway leads to the foot of the tower. A man standing at the top of the tower observes a car at an angle of depression of 30°, which is approaching the foot of the tower at a uniform speed. Six seconds later, the angle of depression of the car is found to be 60°. Find out the time taken by the car to reach the foot of the tower from this point.

Answer 31.

Let’s AB is the tower.

D is the initial, and C is the car’s final position.

Hence man is standing at the top of the tower, so the Angles of depression are measured from A.

BC is the distance from the foot of the tower to the car.

 

Step 1: In the right ΔABC,

tan 60° = AB/BC

√3 = AB/BC

BC = AB/√3

AB = √3 BC

Step 2:

In the right ΔABD,

tan 30° = AB/BD

1/√3 = AB/BD

AB = BD/√3

Step 3: Form Step 1 and Step 2. We have

√3 BC = BD/√3 (Since LHS are the same, so RHS are also the same)

3 BC = BD

3 BC = BC + CD

2BC = CD

or BC = CD/2

Here, the distance of BC is half of CD. Thus, the time taken is also half.

Time taken by car to travel the distance CD = 6 sec. Time taken by car for travel BC = 6/2 = 3 sec.

 

Question 32. The angles of the elevation of the top of a tower from two points at a distance of 4 m and 9 m from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is 6 m.

Answer 32.

Let AB be the tower. C and D are the two points with a distance of 4 m and 9 m from the base, respectively. As per the question,

In the right ΔABC,

tan x = AB/BC

tan x = AB/4

AB = 4 tan x … (i)

Again, from the right ΔABD,

tan (90°-x) = AB/BD

cot x = AB/9

AB = 9 cot x … (ii)

Multiplying equations (i) and (ii)

AB2 = 9 cot x × 4 tan x

⇒ AB2 = 36 (here cot x = 1/tan x

⇒ AB = ± 6

hence height cannot be negative. Therefore, the height of the tower is 6 m.

Hence Proved.