Trigonometry
Class 10th Mathematics Tamilnadu Board Solution
- cos^2 θ + sec^2 θ = 2 + sinθ Determine whether each of the following is an…
- cot^2 θ + cosθ = sin^2 θ Determine whether each of the following is an identity…
- sec^2 θ + cosec^2 θ = sec^2 θ cosec^2 θ Prove the following identities…
- sintegrate heta /1-costheta = cosectheta +cottheta Prove the following…
- root 1-sintegrate heta /1+sintegrate heta = sectheta -tantheta Prove the…
- costheta /sectheta -tantheta = 1+sintegrate heta Prove the following identities…
- root sec^2theta +cosec^2theta = tantheta +cottheta Prove the following…
- 1+costheta -sin^2theta /sintegrate heta (1+costheta) = cottheta Prove the…
- secθ (1 - sinθ)(secθ + tanθ) = 1 Prove the following identities
- sintegrate heta /cosectheta +cottheta = 1-costheta Prove the following…
- sin (90^circle - theta)/1+sintegrate heta + costheta /1-cos (90^circle - theta)…
- tantheta /1-cottheta + cottheta /1-tantheta = 1+sectheta cosectheta Prove the…
- sin (90^circle - theta)/1-tantheta + costheta (90^circle - theta)/1-cottheta =…
- tan (90^circle - theta)/cosectheta +1 + cosectheta +1/cottheta = 2sectheta…
- cottheta +cosectheta -1/cottheta -cosectheta +1 = cosectheta +cottheta Prove…
- (1 + cotθ - cosec θ)(1 + tanθ + secθ) = 2 Prove the following identities.…
- sintegrate heta -costheta +1/sintegrate heta +costheta -1 = 1/sectheta…
- tantheta /1-tan^2theta = sintegrate heta sin (90^circle - theta)/2sin^2…
- 1/cosectheta -cottheta - 1/sintegrate heta = 1/sintegrate heta - 1/cosectheta…
- cot^2theta +sec^2theta /tan^2theta +cosec^2theta = (sintegrate heta costheta)…
- If x = a secθ + b tanθ and y = a tanθ + b secθ, then prove that x^2 - y^2 = a^2…
- If tanθ = n tanα and sinθ = m sinα, then prove that cos^2theta = m^2 - 1/n^2 - 1…
- If sinθ, cosθ and tanθ are in G.P., then prove that cot^6 θ - cot^2 θ = 1.…
- A ramp for unloading a moving truck has an angle of elevation of 30°. If the top…
- A girl of height 150 cm stands in front of a lamp-post and casts a shadow of…
- Suppose two insects A and B can hear each other up to a range of 2 m. The insect…
- To find the cloud ceiling, one night an observer directed a spotlight vertically…
- A simple pendulum of length 40 cm subtends 60° at the vertex in one full…
- Two crows A and B are sitting at a height of 15 m and 10 m in two different…
- A lamp-post stands at the centre of a circular park. Let P and Q be two points…
- A person in an helicopter flying at a height of 700 m, observes two objects…
- A person X standing on a horizontal plane, observes a bird flying at a distance…
- A student sitting in a classroom sees a picture on the black board at a height…
- A boy is standing at some distance from a 30 m tall building and his eye level…
- From the top of a lighthouse of height 200 feet, the lighthouse keeper observes…
- A boy standing on the ground, spots a balloon moving with the wind in a…
- A straight highway leads to the foot of a tower. A man standing on the top of…
- The angles of elevation of an artificial earth satellite is measured from two…
- From the top of a tower of height 60 m, the angles of depression of the top and…
- From the top and foot of a 40 m high tower, the angles of elevation of the top…
- The angle of elevation of a hovering helicopter as seen from a point 45 m above…
- (1 sin^2)sec^2 = Choose the correct answer:A. 0 B. 1 C. tan^2 θ D. cos^2 θ…
- (1 + tan^2)sin^2 = Choose the correct answer:A. sin^2 θ B. cos^2 θ C. tan^2 θ D.…
- (1 cos^2)(1 + cot^2) = Choose the correct answer:A. sin^2 θ B. 0 C. 1 D. tan^2 θ…
- sin(90)cos + cos(90) sin = Choose the correct answer:A. 1 B. 0 C. 2 D. -1…
- 1 - sin^2theta /1+costheta = Choose the correct answer:A. cos θ B. tan θ C. cot…
- cos^4 x sin^4 x = Choose the correct answer:A. 2sin^2 x - 1 B. 2cos^2 x - 1 C. 1…
- If tantheta = a/x , then the value of x/root a^2 + x^2 = Choose the correct…
- If x = a sec, y = b tan , then the value of x^2/a^2 - y^2/b^2 = Choose the…
- sectheta /cottheta +tantheta = Choose the correct answer:A. cot θ B. tan θ C.…
- sin (90^circle - theta) sintegrate heta /tantheta + cos (90^circle - theta)…
- In the adjoining figure, AC = Choose the correct answer:A. 25 m B. 25√3 m C.…
- In the adjoining figure ABC = Choose the correct answer:A. 45° B. 30° C. 60° D.…
- A man is 28.5 m away from a tower. His eye level above the ground is 1.5 m. The…
- In the adjoining figure, sintegrate heta = 15/17 . Then BC = a_s^mathscrn…
- (1 + tan^2)(1 sin)(1 + sin) = Choose the correct answer:A. cos^2 θ - sin^2 θ B.…
- (1 + cot^2)(1 cos)(1 + cos) = Choose the correct answer:A. tan^2 θ - sec^2 θ B.…
- (cos^2 1)(cot^2 + 1) + 1 = Choose the correct answer:A. 1 B. -1 C. 2 D. 0…
- 1+tan^2theta /1+cot^2theta = Choose the correct answer:A. cos^2 θ B. tan^2 θ C.…
- sin^2theta + 1/1+tan^2theta = Choose the correct answer:A. cosec^2 θ + cot^2 θ…
- 9tan^2 9 sec^2 = Choose the correct answer:A. 1 B. 0 C. 9 D. -9
Exercise 7.1
Question 1.Determine whether each of the following is an identity or not.
cos2θ + sec2θ = 2 + sinθ
Answer:
cos2θ + sec2θ = 2 + sinθ
Let θ = 0°
LHS = cos2θ + sec2θ = cos2θ° + sec20° = 1 +1 = 2 … (1)
RHS = 2 + sinθ = 2 + sin0° = 2 + 0 = 2 … (2)
From (1) and (2), LHS = RHS.
∴ The given equation is an identity.
Question 2.
Determine whether each of the following is an identity or not.
cot2θ + cosθ = sin2θ
Answer:
cot2θ + cosθ = sin2θ
Let θ = 45°.
LHS = cot2θ + cosθ = cot245° + cos45° = 1 + 
= 
… (1)
RHS = sin2θ = sin2(45°) = ()2 = 1/2 … (2)
From (1) and (2), LHS ≠ RHS.
∴ The given equation is not an identity.
Question 3.
Prove the following identities
sec2θ + cosec2θ = sec2θ cosec2θ
Answer:
Consider LHS,
LHS = sec2θ + cosec2θ
We know that secθ = 
and cosecθ = 
⇒ sec2θ + cosec2θ = 
+ 
= 
We know that sin2θ + cos2θ = 1.
⇒ = 
= ×
= sec2θ cosec2θ = RHS
Hence proved.
Question 4.
Prove the following identities
Answer:
Consider LHS,
LHS = 
Multiplying numerator and denominator by (1 + cosθ),
⇒ = × 
We know that (a + b) (a – b) = a2 – b2
⇒ = 
We know that 1 – cos2θ = sin2θ.
⇒ = 
= 
+ 
= + 
We know that = cosecθ and = cotθ
∴ = cosecθ + cotθ = RHS
Hence proved.
Question 5.
Prove the following identities
Answer:
Consider LHS,
LHS = 
Multiplying and dividing with (1 – sinθ),
⇒ = 
We know that (a + b) (a – b) = a2 – b2
⇒ = 
We know that 1 – sin2θ = cos2θ.
⇒ = 
= 
= 
We know that = secθ and 
= tanθ.
∴ = secθ + tanθ = RHS
Hence proved.
Question 6.
Prove the following identities
Answer:
Consider LHS,
LHS = 
Multiplying and dividing with (secθ + tanθ),
⇒ = × 
We know that (a + b) (a – b) = a2 – b2.
⇒ = 
We know that sec2θ - tan2θ = 1.
⇒ = 
We know that = secθ and = tanθ.
⇒ = cosθ() + cosθ ()
= 1 + sinθ = RHS
Hence proved.
Question 7.
Prove the following identities
Answer:
Consider LHS,
LHS = 
We know that sec2θ = 1 + tan2θ and coesc2θ = 1 + cot2θ.
⇒ = 
= 
= 
We know that tanθ cotθ = 1.
⇒ =
We know that a2 + b2 + 2ab = (a + b)2.
⇒ =
= (tanθ + cotθ) = RHS
Hence proved.
Question 8.
Prove the following identities
Answer:
Consider LHS,
LHS = 
We know that sin2θ = 1 – cos2θ.
⇒ = 
= 
= 
=
We know that = cotθ.
∴ = cotθ =RHS
Hence proved.
Question 9.
Prove the following identities
secθ (1 – sinθ)(secθ + tanθ) = 1
Answer:
Consider LHS,
LHS = secθ (1 – sinθ) (secθ + tanθ)
⇒ secθ (1 – sinθ) (secθ + tanθ) = (secθ - ) (secθ + tanθ)
We know that = tanθ.
⇒ secθ (1 – sinθ) (secθ + tanθ) = (secθ – tanθ) (secθ + tanθ)
We know that (a + b) (a – b) = a2 – b2.
⇒ secθ (1 – sinθ) (secθ + tanθ) = sec2θ – tan2θ
We know that sec2θ – tan2θ = 1.
∴ secθ (1 – sinθ) (secθ + tanθ) = 1 = RHS
Hence proved.
Question 10.
Prove the following identities
Answer:
Consider LHS,
LHS = 
Multiplying and dividing with (cosecθ - cotθ),
⇒ = × 
We know that (a + b) (a – b) = a2 – b2.
⇒ = 
We know that cosec2θ - cot2θ = 1.
⇒ = 
We know that = cosecθ and = cotθ.
⇒ = sinθ () - sinθ ()
= 1 – cosθ = RHS
Hence proved.
Question 11.
Prove the following identities.
Answer:
Consider LHS,
LHS = 
+ 
We know that sin (90° - θ) = cosθ and cos (90° - θ) = sinθ.
⇒ + = 
+ 
Taking LCM,
⇒ + = 
We know that (a + b) (a – b) = a2 – b2.
⇒ + = 
We know that 1 –sin2θ = cos2θ.
⇒ + = 
= 2 ()
We know that = secθ.
∴ + = 2secθ = RHS
Hence proved.
Question 12.
Prove the following identities.
Answer:
Consider LHS,
LHS = 
+ 
We know that = cotθ and = tanθ.
⇒ + = 
+ 
= 
+ 
= 
+ 
= 
[
- 
]
= [
]
We know that a3 – b3 = (a – b) (a2 + ab + b2).
⇒ + = [
]
We know that sin2θ + cos2θ = 1.
⇒ + = 
= 
+ 
We know that = cosecθ and = secθ.
∴ ⇒ + = 1 + secθcosecθ = RHS
Hence proved.
Question 13.
Prove the following identities.
Answer:
Consider LHS,
LHS = 
+ 
We know that sin (90° - θ) = cosθ and cos (90° - θ) = sinθ.
⇒ + = 
+ 
We know that = cotθ and = tanθ.
⇒ + = 
+ 
= 
+ 
= 
- 
= 
We know that (a + b) (a – b) = a2 – b2.
⇒ + = 
= cosθ + sinθ = RHS
Hence proved.
Question 14.
Prove the following identities.
Answer:
Consider LHS,
LHS = 
+ 
We know that tan (90° - θ) = cotθ.
⇒ + = 
+
= 
We know that (a + b)2 = a2 + 2ab + b2.
⇒ + = 
We know that cot2θ + 1 = cosec2θ.
⇒ + = 
= 
= 
= 
We know that = cosecθ and = cotθ.
⇒ + = 
= 
We know that = secθ.
⇒ + = 2secθ = RHS
Hence proved.
Question 15.
Prove the following identities.
Answer:
Consider LHS,
LHS = 
We know that cosec2θ – cot2θ = 1.
⇒ = 
We know that (a + b) (a – b) = a2 – b2.
⇒ = 
= 
= 
= cosecθ + cotθ = RHS
Hence proved.
Question 16.
Prove the following identities.
(1 + cotθ – cosec θ)(1 + tanθ + secθ) = 2
Answer:
Consider LHS,
LHS = (1 + cotθ – cosecθ) (1 + tanθ + secθ)
Expanding the above,
⇒ (1 + cotθ – cosecθ) (1 + tanθ + secθ)
= 1 + tanθ + secθ + cotθ + cotθtanθ + cotθsecθ – cosecθ –cosecθtanθ –cosecθsecθ
We know that = secθ,
= cosecθ,
= tanθ and = cotθ.
= 1 + tanθ + secθ + cotθ + 1 + cosecθ – cosecθ – secθ – cosecθsecθ
We know that tanθ + cotθ = cosecθsecθ.
= 1 + cosecθsecθ – cosecθsecθ + 1
∴ (1 + cotθ – cosecθ) (1 + tanθ + secθ) = 2 = RHS
Hence proved.
Question 17.
Prove the following identities.
Answer:
Consider LHS,
LHS = 
Multiplying and dividing by sinθ + cosθ + 1,
⇒ = × 
We know that (a + b) (a – b) = a2 – b2.
⇒ = 
= 
We know that 1 – cos2θ = sin2θ and sin2θ + cos2θ = 1.
= 
= 
= 
= 
We know that = secθ and = tanθ.
⇒ = secθ + tanθ
Multiplying and dividing by secθ – tanθ,
⇒ = secθ + tanθ × 
= 
We know that sec2 – tan2θ = 1.
∴ =
= RHS
Hence proved.
Question 18.
Prove the following identities.
Answer:
Consider RHS,
RHS = 
We know that sin (90° - θ) = cosθ.
⇒ = 
We know that 2cos2θ – 1 = cos2θ – sin2θ
⇒ = 
Multiplying and dividing by cos2θ,
⇒ = 
= 
= LHS
Hence proved.
Question 19.
Prove the following identities.
Answer:
Consider LHS,
LHS = 
-
We know that cosec2θ – cot2θ = 1 and = cosecθ.
⇒ - = 
– cosecθ
We know that (a + b) (a – b) = a2 – b2.
⇒ - = 
– cosecθ
= cosecθ + cotθ – cosecθ
= cotθ … (1)
Now consider RHS,
RHS = - 
We know that cosec2θ – cot2θ = 1 and = cosecθ.
⇒ - = cosecθ - 
We know that (a + b) (a – b) = a2 – b2.
⇒ - =cosecθ - 
= cosecθ – (cosecθ – cotθ)
= cotθ … (2)
From (1) and (2), LHS = RHS
Hence proved.
Question 20.
Prove the following identities.
Answer:
Consider LHS,
LHS = 
We know that cot2θ = cosec2 – 1 and tan2θ = sec2θ – 1.
⇒ = 
= 
= 1 … (1)
Consider RHS,
RHS = (sinθcosθ) (tanθ + cotθ)
Expanding,
⇒ (sinθcosθ) (tanθ + cotθ) = sinθcosθtanθ + sinθcosθcotθ
We know that = tanθ and = cotθ.
⇒ (sinθcosθ) (tanθ + cotθ) = sinθcosθ() + sinθcosθ()
= sin2θ + cos2θ
We know that sin2θ + cos2θ = 1.
∴ (sinθcosθ) (tanθ + cotθ) = 1 … (2)
From (1) and (2), LHS = RHS.
Hence proved.
Question 21.
If x = a secθ + b tanθ and y = a tanθ + b secθ, then prove that x2 – y2 = a2 – b2.
Answer:
x2 =(a secθ + b tanθ)2
=a2sec2θ + b2tan2θ + ab secθ tanθ
y2 =(a tanθ + b secθ)2
=a2tan2θ + b2sec2θ + ab secθ tanθ
Now,
LHS = x2– y2
= (a2sec2θ + b2tan2θ + ab secθ tanθ)
- (a2tan2θ + b2sec2θ + ab secθ tanθ)
= a2sec2θ + b2tan2θ + ab secθ tanθ
- a2tan2θ - b2sec2θ - ab secθ tanθ
= a2 (sec2θ –tan2θ) – b2 (sec2θ –tan2θ)
=a2 – b2 [∵ sec2θ –tan2θ =1]
=RHS
Hence proved.
Question 22.
If tanθ = n tanα and sinθ = m sinα, then prove that 
Answer:
We want to find value of cos2θ in terms of m and n.
So we first eliminate angle α,
tanθ =n tanα [∵ Given]
⇒ 
⇒ 
(1)
sinθ =m sinα [∵ Given]
⇒ 
⇒ 
(2)
We know that,
cosec2α – cot2α =1
Substituting values from (1) and (2) gives,
Hence proved.
Question 23.
If sinθ, cosθ and tanθ are in G.P., then prove that cot6θ – cot2θ = 1.
Answer:
Given: sinθ, cosθ, tanθ are in G.P.
So,
cos2θ =sinθ × tanθ
cos2θ =sinθ ×
cos2θ =
Or, 
⇒ cot2θ =secθ (1)
Taking LHS= cot6θ –cot2θ
= (cot2θ) 3 – cot2θ
=sec3θ – secθ [Substituting from eqn. (1)]
=secθ (sec2θ -1)
=secθ (tan2θ)
=cot2θ.tan2θ [Substituting from eqn. (1)]
=1
=RHS
Hence proved.
Exercise 7.2
Question 1.A ramp for unloading a moving truck has an angle of elevation of 30°. If the top of the ramp is 0.9 m above the ground level, then find the length of the ramp.
Answer:
Given AB = 0.9 m, AC = ?
In the above right triangle,
AC = Hypotenuse
AB = Perpendicular
BC = Base
We know,
⇒ AC = 0.9 x 2
⇒ AC = 1.8 m
∴ length of ramp is 1.8 m
Question 2.
A girl of height 150 cm stands in front of a lamp–post and casts a shadow of length 150√3 cm on the ground. Find the angle of elevation of the top of the lamp–post.
Answer:
Given, AB = 150cm, BC = 150√3 and ∠C = ?
Here, AC = Hypotenuse
AB = Perpendicular
BC = Base
We know,
⇒ tan θ = tan 30°
⇒ θ = 30°
∴ the angle of elevation of the top of the lamp–post is 30°
Question 3.
Suppose two insects A and B can hear each other up to a range of 2 m. The insect A is on the ground 1 m away from a wall and sees her friend B on the wall, about to be eaten by a spider. If A sounds a warning to B and if the angle of elevation of B from A is 30°, will the spider have a meal or not? ( Assume that B escapes if she hears A calling)
Answer:
Given, BC = 1 m and A and B can hear each other up to a range of 2 m.
Here, AC = Hypotenuse
BC = Perpendicular
AC = Base
We know,
⇒ AC = 2 m
⇒ A and B can hear each other.
∴ Insect B escapes.
⇒ Spider will not have a meal.
Question 4.
To find the cloud ceiling, one night an observer directed a spotlight vertically at the clouds. Using a theodolite placed 100 m from the spotlight and 1.5 m above the ground, he found the angle of elevation to be 60°. How high was the cloud ceiling? (Hint : See figure)
(Note: Cloud ceiling is the lowest altitude at which solid cloud is present. The cloud ceiling at airports must be sufficiently high for safe take offs and landings. At night the cloud ceiling can be determined by illuminating the base of the clouds by a spotlight pointing vertically upward.)
Answer:
Given, Base = 100 m and height of cloud ceiling = ?
We know,
⇒ 100 √3 = AB
⇒ AB = 100 √3
= 100 (1.732)
= 173.2
⇒ height of ceiling from ground = 173.2 + 1.5
= 174.7 m
Question 5.
A simple pendulum of length 40 cm subtends 60° at the vertex in one full oscillation. What will be the shortest distance between the initial position and the final position of the bob? (between the extreme ends)
Answer:
Given, OA = OC = length of pendulum = 40 cm, ∠AOC = 60°
In triangle OBC,
= 30°
Here, OC = hypotenuse
AB = perpendicular
BC = base
We know,
⇒ 40 = 2 BC
⇒ BC = 20 cm
∴ length of AC = 2(BC)
= 2(20)
= 40 cm
∴ the shortest distance between the initial position and the final position of the bob = 40 cm
Question 6.
Two crows A and B are sitting at a height of 15 m and 10 m in two different trees vertically opposite to each other . They view a vadai (an eatable) on the ground at an angle of depression 45° and 60° respectively. They start at the same time and fly at the same speed along the shortest path to pick up the vadai. Which bird will succeed in it? Hint : (foot of two trees and vadai (an eatable) are in a straight line)
Answer:
Given, AC = 15 m, BD = 10 m, AE = ? and BE = ?
In triangle BED,
BE = hypotenuse
BD = perpendicular
ED = Base
∠ BED = ∠ OBE (adjacent angles are equal)
= 60°
We know,
⇒ BE√3 = 10 x 2
= 11.55 m
And, in triangle AEC ,
AC = Perpendicular
AE = Hypotenuse
CE = Base
∠ AEC = ∠ MAE (adjacent angles are equal)
= 45°
We know,
⇒ AE = 15√2
= 21.21 m
⇒ BD<AE
⇒ Crow B will succeed.
Question 7.
A lamp–post stands at the centre of a circular park. Let P and Q be two points on the boundary such that PQ subtends an angle 90° at the foot of the lamp–post and the angle of elevation of the top of the lamp post from P is 30°. If PQ = 30 m, then find the height of the lamp post.
Answer:
Given that PQ = 30m and ∠POQ = 90°
Let O be the centre of the park and OR be the lamp post and P and Q be two points on the boundary of the circular park.
In a right triangle OPQ,
OP = OQ = radius.
⇒ ∠OPQ = ∠OQP = 45° (∵ POQ = 90°)
We know,
Multiplying and dividing ihe fraction by √2,we get–
= 15 √2
And, in triangle RPO,
Multiplying and dividing the fraction by √6, we get–
⇒ OR = 5√6
∴ the height of the lamp post is 5√6 m
Question 8.
A person in an helicopter flying at a height of 700 m, observes two objects lying opposite to each other on either bank of a river. The angles of depression of the objects are 30° and 45°. Find the width of the river. (√3 = 1.732 )
Answer:
Given, AD = 700 m and BC = ?
In triangle ACD,
∠ACD = ∠ MAC (alternate angles are equal)
= 45°
We know,
⇒ DC = 700 m …… (1)
In triangle ABD,
∠ABD = ∠ OAB
= 30° (alternate angles are equal)
We know,
⇒ BD = 700√3 …… (2)
Now, adding equation (1) and (2)–
Width of the river = BD + DC
= 700 + 700√3
= 700(1 + √3)
= 700 (1 + 1.732)
= 700 × 2.732
= 1912.40 m
∴ Width of the river is 1912.40 m
Question 9.
A person X standing on a horizontal plane, observes a bird flying at a distance of 100 m from him at an angle of elevation of 30c. Another person Y standing on the roof of a 20 m high building, observes the bird at the same time at an angle of elevation of 45°. If X and Y are on the opposite sides of the bird, then find the distance of the bird from Y.
Answer:
Given, AC = 700 m and EF = ?
Let position of person X be B, position of person Y be F and AE = x.
In triangle ABC
∠ABC = 30°
We know,
⇒ 100 = 2 (x + 20)
⇒ 100 = 2x + 40
⇒ 2x = 60
⇒ x = 30
And, in triangle AFE,
∠AFE = 45°
⇒ EF = 30√2
∴ Distance of the bird from Y is 30√2 m.
Question 10.
A student sitting in a classroom sees a picture on the black board at a height of 1.5 m from the horizontal level of sight. The angle of elevation of the picture is 30°. As the picture is not clear to him, he moves straight towards the black board and sees the picture at an angle of elevation of 45°. Find the distance moved by the student.
Answer:
Given, AD = 1.5 m and distance moved = BC = ?
In triangle ABD,
∠ABD = 30°
We know,
⇒ BD = 1.5 x √3
⇒ BD = 1.5√3
Now, in triangle ACD,
∠ACD = 45°
We know,
⇒ CD = 1.5
⇒ BC = BD – CD
⇒ BC = 1.5 √3 – 1.5
⇒ BC = 1.5 (√3 – 1)
⇒ BC = 1.5(1.732 – 1)
⇒ BC = 1.5 (0.732)
⇒ BC = 1.098 m
∴ the distance moved by the student is 1.098 m.
Question 11.
A boy is standing at some distance from a 30 m tall building and his eye level from the ground is 1.5 m. The angle of elevation from his eyes to the top of the building increases from 30° to 60° as he walks towards the building. Find the distance he walked towards the building.
Answer:
Given, AF = 30m, DF = BE = 1.5m,
AD = 28.5 m
In triangle ABD
∠ABD = 30°
We know,
⇒ BD = 28.5 √3
Now, in triangle ACD,
∠ABD = 60°
Multiplying and dividing the fraction by √3, we get
⇒ CD = 9.5√3
∴ distance he walked towards the building = BD – CD
= 28.5√3 –9.5√3
= 19√3 m
Question 12.
From the top of a lighthouse of height 200 feet, the lighthouse keeper observes a Yacht and a Barge along the same line of sight . The angles of depression for the Yacht and the Barge are 45° and 30° respectively. For safety purposes the two sea vessels should be atleast 300 feet apart. If they are less than 300 feet, the keeper has to sound the alarm. Does the keeper have to sound the alarm?
Answer:
Given, AB = 200 ft. and CD = ?
In triangle ABC,
∠ACB = ∠ OAC (alternate angles are equal)
= 45°
We know,
⇒ BC = 200 √2
And, in triangle ABD,
∠ADB = ∠OAD (alternate angles are equal)
= 30°
⇒ BD = 200 x 2
⇒ BD = 400
Now, CD = BD – BC
⇒ CD = 400 – 200 √2
= 200(2 – √2)
= 200 (2 – 1.414)
= 200(0.586)
= 117.2 m
∴distance between Yacht and a Barge = 117.2<300 m.
⇒ the keeper has to sound the alarm.
Question 13.
A boy standing on the ground, spots a balloon moving with the wind in a horizontal line at a constant height. The angle of elevation of the balloon from the boy at an instant is 60°. After 2 minutes, from the same point of observation, the angle of elevation reduces to 30°. If the speed of wind is 29√3 m/min. then, find the height of the balloon from the ground level.
Answer:
Here, Distance covered by the balloon = BC
We know,
Distance = Time x Speed
⇒ BC = Time x Speed
= 2 x 29√3
= 58√3 m
Let AB = x
⇒ AC = x + 58√3
In triangle DAC,
∠DAC = 30°
We know,
Now, in triangle EAB,
∠EAB = 60°
⇒ EB = √3x
∵ EB = DC
⇒ x + 58√3 = 3x
⇒ 2x = 58√3
⇒ x = 29√3 m
And, Height of the balloon from ground level EB = √3 x
= 29 √3 (√3)
= 87 m
Hence height of the balloon from ground level is 87 m.
Question 14.
A straight highway leads to the foot of a tower. A man standing on the top of the tower spots a van at an angle of depression of 30°. The van is approaching the tower with a uniform speed. After 6 minutes, the angle of depression of the van is found to be 60°. How many more minutes will it take for the van to reach the tower?
Answer:
Given, time taken by van to reach D from C = 6 minutes.
And let the speed = x
We know,
Distance = speed × time
⇒ Distance between D and C = DC = 6x
In triangle ACB,
∠ACB = ∠ OAC (alternate angles are equal)
= 30°
Also,
……… (1)
Now, in triangle ABD,
∠ABD = ∠ OAD (alternate angles are equal)
= 60°
We know,
AB = BD√3 ……… (2)
Now, equating (1) & (2), we get–
⇒ BD + 6x = BD × 3
⇒ 2BD = 6x
⇒ BD = 3x (where, x is speed)
Now, comparing it with Distance = speed × time, we have–
Time = 3 minutes.
Hence, it take 3 minutes more for the van to reach the tower.
Question 15.
The angles of elevation of an artificial earth satellite is measured from two earth stations, situated on the same side of the satellite, are found to be 30° and 60°. The two earth stations and the satellite are in the same vertical plane. If the distance between the earth stations is 4000 km, find the distance between the satellite and earth. (√3 = 1.732)
Answer:
Let C be position of station 1 and D of station 2 and BC = x.
Given, CD = 4000 km
Now, in triangle ABC,
∠ACB = 60°
We know,
⇒ AB = √3BC
⇒ AB = x√3 ……… (1)
And, in triangle ABD,
∠ADB = 30°
We know,
……… (2)
Now, equating (1) & (2), we get–
⇒ 3 x = x + 4000
⇒ 3x – x = 4000
⇒ 2x = 4000
⇒ x = 2000
And distance between the satellite and earth(AB) = x√3
= 2000(1.732)
= 3464 km
∴ Distance between the satellite and earth = 3464 km
Question 16.
From the top of a tower of height 60 m, the angles of depression of the top and the bottom of a building are observed to be 30° and 60° respectively. Find the height of the building.
Answer:
Here, AB is tower and EC is building.
Given, AB = 60 m and EC = ?
In triangle ABC,
Let AB = x
⇒ BD = 60 – x
And, ∠ ACB = ∠OAC (alternate angles are equal)
∠ACB = 30°
We know,
⇒ BC = x √3 ……… (1)
In triangle ADE,
∠ AED = ∠OAE (alternate angles are equal)
∠AED = 60°
……… (2)
∵ BC = DE
⇒ equation (1) = equation (2)
⇒ 3x = 60
⇒ x = 20 m
∵ CE = BD and BD = 60–x.
⇒ CE = 60 – 20 = 40 m
∴ Height of the building = 40 m.
Question 17.
From the top and foot of a 40 m high tower, the angles of elevation of the top of a lighthouse are found to be 30° and 60° respectively. Find the height of the lighthouse. Also find the distance of the top of the lighthouse from the foot of the tower.
Answer:
Given, CE = 40 m
Let AB = x and BD = CE = 40 m
In triangle ABC ,
∠ACB = 30°
We know,
⇒ BC = x√3 ……… (1)
And, in triangle ADE,
∠AED = 60°
We know,
……… (2)
∵ BC = DE
⇒ equation (1) = equation (2)
⇒ 3 x = x + 40
⇒ 2x = 40
⇒ x = 20
∴ Height of the tower = 40 + x
= 40 + 20
= 60 m
And, from (2)–
DE = BC
= x√3
= 20√3
And, in triangle ADE,
Also we know,
⇒ AE = 40√3 m
∴ the height of the lighthouse is 60 m and the distance of the top of the lighthouse from the foot of the tower is 40√3 m.
Question 18.
The angle of elevation of a hovering helicopter as seen from a point 45 m above a lake is 30° and the angle of depression of its reflection in the lake, as seen from the same point and at the same time, is 60°. Find the distance of the helicopter from the surface of the lake.
Answer:
Given, ED = 45 m and DF = ?
Let EF = x, DF = h
⇒ DC = h (∵ height of reflection = height of object)
In a right triangle FAE,
We know,
AE = (h – 45) √3 ……… (1)
And, in right triangle ACE,
………(2)
Now, ∵BC = AD
⇒ equation (1) = equation (2)
⇒ 3(h – 45) = 45 + h
⇒ 3h – 135 = 45 + h
⇒ 2h = 45 + 135
⇒ 2h = 180
⇒ h = 90 m
∴ the distance of the helicopter from the surface of the lake is
90 m.
Exercise 7.3
Question 1.Choose the correct answer:
(1 – sin2θ)sec2θ =
A. 0
B. 1
C. tan2θ
D. cos2θ
Answer:
Given trigonometric equation = 
We know that,
∴ 
⇒ (1–
)
= 
⇒ (1–) = 
(∵ 
)
⇒ (1–) = 1
Hence, (1–) = 1.
Question 2.
Choose the correct answer:
(1 + tan2θ)sin2θ =
A. sin2θ
B. cos2θ
C. tan2θ
D. cot2θ
Answer:
Given Trigonometric equation = 
We know that,
∴ = 
⇒ = 
⇒ = 
(∵ )
= 
= 
Hence, = .
Question 3.
Choose the correct answer:
(1 – cos2θ)(1 + cot2θ) =
A. sin2θ
B. 0
C. 1
D. tan2θ
Answer:
Given trigonometric equation is = 
As we know that,
∴ 
And 
⇒ 
⇒ 
⇒ 
⇒ 
Hence, ⇒ 
Question 4.
Choose the correct answer:
sin(90° – θ)cosθ + cos(90° – θ) sin θ =
A. 1
B. 0
C. 2
D. –1
Answer:
From Trigonometric identities –
⇒ 
⇒ 
Using above formulas.
⇒ 
⇒ 
⇒ 
(∵ )
Question 5.
Choose the correct answer:
=
A. cos θ
B. tan θ
C. cot θ
D. cosec θ
Answer:
Given, 
⇒ 
[taking (1+cosθ) as L.C.M.)
⇒ 
(∵ sin2θ = 1 – cos2θ)
⇒ 
⇒ 
⇒ 
⇒ 
Hence,
Question 6.
Choose the correct answer:
cos4x – sin4x =
A. 2sin2 x – 1
B. 2cos2 x – 1
C. 1 + 2 sin2x
D. 1 – 2 cos2x
Answer:
we can write –
∴ 
Now applying formula,
⇒ 
We know that, 
⇒ 
⇒ 
(∵ 
⇒ 
⇒ 
Hence,
Question 7.
Choose the correct answer:
If 
, then the value of 
=
A. cos θ
B. sin θ
C. cosec θ
D. sec θ
Answer:
Given, 
∴ a=x tanθ
⇒ 
⇒ 
⇒ 
⇒ 
(∵ sec2θ – tan2θ =1)
⇒ 
⇒ 
(∵ 
Question 8.
Choose the correct answer:
If x = a secθ, y = b tan θ , then the value of 
=
A. 1
B. –1
C. tan2θ
D. cosec2θ
Answer:
Given,
⇒ 
⇒ 
⇒ 
⇒ 
(∵ 
)
Question 9.
Choose the correct answer:
=
A. cot θ
B. tan θ
C. sin θ
D. – cot θ
Answer:
We know that,
⇒ 
⇒ 
⇒ 
(∵ cos2θ +sin2θ=1)
⇒ 
⇒ 
Hence, proved 
Question 10.
Choose the correct answer:
=
A. tan θ
B. 1
C. –1
D. sin θ
Answer:
We know that,
⇒ 
⇒ 
⇒ 
⇒ 
⇒ 
(∵ 
)
Question 11.
Choose the correct answer:
In the adjoining figure, AC =
A. 25 m
B. 25√3 m
C. 
m
D. 25√2 m
Answer:
As we know that,
From given figure,
AB = 25 m.
∴ 
⇒ 
⇒ 
(∵ 
)
Question 12.
Choose the correct answer:
In the adjoining figure ∠ABC =
A. 45°
B. 30°
C. 60°
D. 50°
Answer:
From given figure,
⇒ 
⇒ 
⇒ 
⇒ 
⇒ 
Question 13.
Choose the correct answer:
A man is 28.5 m away from a tower. His eye level above the ground is 1.5 m. The angle of elevation of the tower from his eyes is 45c. Then the height of the tower is
A. 30 m
B. 27.5 m
C. 28.5 m
D. 27 m
Answer:
figure can be drawn as –
∠CDE=45°
⇒
⇒ 
Height of the tower = 28.5+1.5= 30 m
Question 14.
Choose the correct answer:
In the adjoining figure, 
. Then BC =
A. 85 m
B. 65 m
C. 95 m
D. 75 m
Answer:
We know that,
⇒ 
Given that, 
⇒ 
(from figure AC = 85 m)
⇒ 
Question 15.
Choose the correct answer:
(1 + tan2θ)(1 – sinθ)(1 + sinθ) =
A. cos2θ – sin2θ
B. sin2θ – cos2θ
C. sin2θ + cos2θ
D. 0
Answer:
We know that,
⇒ 
Using formula, 
⇒ 
⇒ 
(∵ )
⇒ 
Hence,
Question 16.
Choose the correct answer:
(1 + cot2θ)(1 – cosθ)(1 + cos θ) =
A. tan2θ – sec2θ
B. sin2θ – cos2θ
C. sec2θ – tan2θ
D. cos2θ – sin2θ
Answer:
We know that,
And 
⇒ 
[∵ 
⇒ 
∵ 
∴ 
⇒ 
Also, we know that,
∴
Question 17.
Choose the correct answer:
(cos2θ – 1)(cot2θ + 1) + 1 =
A. 1
B. –1
C. 2
D. 0
Answer:
We know that,
Now,
⇒ 
∵ 
And
⇒ 
⇒ 
⇒ 
Hence, 
Question 18.
Choose the correct answer:
=
A. cos2θ
B. tan2θ
C. sin2θ
D. cot2θ
Answer:
We know that,
∴ 
Where, 
and 
Hence, 
Question 19.
Choose the correct answer:
=
A. cosec2θ + cot2θ
B. cosec2θ – cot2θ
C. cot2θ – cosec2θ
D. sin2θ – cos2θ
Answer:
We know that,
∴
⇒ 
⇒ 
(∵ 
)
⇒ 
(∵ )
Also,
Question 20.
Choose the correct answer:
9tan2θ – 9 sec2θ =
A. 1
B. 0
C. 9
D. –9
Answer:
We know that,
Now, from given questions –
⇒ 
⇒ 
[from equation (i)]
⇒ 