NCERT Exemplar Class 10 Maths Solutions Chapter 8 Introduction To Trigonometry and Its Applications

CBSETuts.com provides you Free PDF download of NCERT Exemplar Class 10 Maths chapter 8 Introduction to Trigonometry and its Applications solved by expert teachers as per NCERT (CBSE) book guidelines. All the chapterwise questions with solutions to help you to revise complete CBSE syllabus and score more marks in Your board examinations. According to new CBSE Exam Pattern, MCQ Questions for Class 10 Maths Carries 20 Marks.

You can also Download NCERT Solutions for class 10 Maths in Hindi to help you to revise complete Syllabus and score more marks in your examinations.

NCERT Exemplar Problems Class 10 Maths Solutions Chapter 8 Introduction To Trigonometry and Its Applications

Exercise 8.1 Multiple Choice Questions (MCQs)

Question 1:
If cos A =\(\frac { 4 }{ 5 } \), then the value of tan A is
(a) \(\frac {3 }{ 5 }\)
(b) \(\frac {3 }{ 4 }\)
(c) \(\frac {4 }{ 3 }\)
(d) \(\frac {5}{ 3 }\)
Solution:

Question 2:
If sin A =\(\frac {1 }{ 2 }\)  then the value of cot A is
(a) √3
(b) \(\frac { 1 }{ \sqrt { 3 }}\)
(c) \(\frac {\sqrt { 3 }}{ 1}\)
(d) 1
Solution:

Question 3:
The value of the expression cosec (75° + 0) – sec (15° – 0) – tan (55° + 0) + cot (35° – 0) is
(a) -1                          (b) 0                            (c)1                            (d) \(\frac {3 }{ 2 }\)
Solution:
(b) Given, expression = cosec (75° + 0) – sec (15° – 0) – tan (55° + 0) + cot (35° – 0)
=cosec [90° – (15° – 0)] – sec (15° – 0)- tan (55° + 0) + cot (90° – (55° + 0)}
= sec (15° – 0) – sec (15° – 0) – tan (55° + 0) + tan (55° + 0)
[∴ cosec (90° – 0) = sec 0 and cot (90° – 0) = tan 0]
= 0
Hence, the required value of the given expression is 0.

Question 4:
If sinθ =\(\frac {3 }{ 5 }\), then cosθ is equal to

Solution:

Question 5:
If cos(α+ β) = 0, then sin (α – β) can be reduced to
(a) cos β                         (b) cos 2β                                  (c) sin α                                    (d) sin 2α
Solution:

Question 6:
The value of (tanl° tan2° tan3°… tan89°) is
(a) 0                           (b) 1                           (c) 2                            (d) \(\frac {1 }{ 2 }\)
Solution:
(b) tan1°-tan2°-tan3°… tan 89°
= tan1°-tan2°-tan3°… tan44° . tan 45° . tan 46°… tan 87°-tan 88°tan 89°
= tan 1°- tan2 °- tan 3°… tan 44° . (1)- tan (90° – 44°)… tan (90° – 3°)
tan (90° -2°)- tan (90° -1°) (∴ tan 45° = 1)
= tan1°-tan2°-tan3°…. tan44° (1) . cot 44°……. cot3°-cot2°-cot1°

Question 7:
If cos 9α = sin α and 9α < 90° ,then the value of tan 5α is
(a) \(\frac { 1 }{ \sqrt { 3 }}\)
(b) \(\frac {\sqrt { 3 }}{ 1}\)
(c) 1
(d) 0
Solution:

Question 8:
If ΔABC is right angled at C, then the value of cos (A + B) is
(a) 0
(b) 1
(c) \(\frac { 1 }{ 2 } \)
(d) \(\frac {\sqrt { 3 }}{ 2}\)
Solution:

Question 9:
If sin A + sin² A = 1, then the value of (cos² A + cos⁴ A) is
(a) 1                      (b) \(\frac { 1 }{ 2 } \)                        (c) 2                                          (d) 3
Solution:

Question 10:
If sinα = \(\frac { 1 }{ 2 } \) and cosβ = \(\frac { 1 }{ 2 } \) then the value of (α + β) is
(a) 0°                (b) 30°                        (c) 60°                     (d) 90°
Solution:

Question 11:
The value of the expression

(a) 3                                 (b) 2                                        (c) 1                              (d) 0
Solution:

Question 12:

(a) \(\frac { 2 }{ 3 } \)
(b) \(\frac { 1 }{ 3 } \)
(c) \(\frac { 1 }{ 2 } \)
(d) \(\frac { 3 }{ 4 } \)
Solution:

Question 13:
If sinθ – cosθ= 0, then the value of (sin⁴ θ + cos⁴ θ) is
(a) 1
(b) \(\frac { 3 }{ 4 } \)
(c) \(\frac { 1 }{ 2 } \)
(d) \(\frac { 1 }{ 4 } \)
Solution:

Question 14:
sin (45° + θ) – cos (45° – θ) is equal to
(a) 2 cosθ                  (b) 0                            (c) 2 sinθ                   (d) 1
Solution:
(b) sin(45° + θ) – cos(45° – θ) = cos[90°- (45° + θ)] – cos(45°- 6) [∴ cos(90° – θ) = sin0]
= cos (45° – 0) – cos (45° – 0)
= 0

Question 15:
If a pole 6 m high casts a shadow 2√3 m long on the ground, then the Sun’s elevation is
(a) 60°                (b) 45°                        (c) 30°                     (d) 90°
Solution:

Exercise 8.2 Very Short Answer Type Questions

Write whether True or False and justify your answer.

Question 1:

Solution:
True

Question 2:
The value of the expression (cos² 23° – sin² 67°) is positive.
Solution:
False
cos²23° – sin² 67°= (cos23° – sin67°)(cos23° + sin67°) [∴(a² – b²) = (a – b) (a + b)]
= [cos23° – sin (90° – 23°)] (cos23° + sin67°)
= (cos23° – cos23°) (cos23° + sin67°) [∴sin (90° – 0) = cos 0]
= 0.(cos23° + sin67°) = 0
which may be either positive or negative.

Question 3:
The value of the expression (sin 80° – cos 80°) is negative.
Solution:
False
We know that, sine is increasing when, O°<θ<9O° and cosθ is decreasing when,
O°<θ<9O°.
∴                                               sin80°-cos80°> 0                                                      [positive]

Question 4:

Solution:
True

Question 5:
If cosA + cos² A = 1, then sin² A + sin⁴ A = 1
Solution:
True

Question 6:
(tanθ + 2) (2 tanθ + 1) = 5 tanθ + sec² θ
Solution:
False

Question 7:
If the length of the shadow of a tower is increasing, then the angle of elevation of the Sun is also increasing.
Solution:
False

Hence, we conclude from above two examples that if the length of the shadow of a tower is increasing, then the angle of elevation of the Sun is decreasing.
Alternate Method
False, we know that, if the elevation moves towards the tower, it increases and if its elevation moves away the tower, it decreases. Hence, if the shadow of a tower is increasing, then the angle of elevation of a Sun is not increasing.

Question 8:
If a man standing on a plat form 3 m above the surface of a lake observes a cloud and its reflection in the lake, then the angle of elevation of the cloud is equal to the angle of depression of its reflection.
Solution:
False
From figure, we observe that, a man standing on a platform at point P, 3 m above the surface of a lake observes a cloud at point C. Let the height of the cloud from the surface of the platform is h and angle of elevation of the cloud is \({\theta}_{1}\).
Now at same point P a man observes a cloud reflection in the lake at this time the height of reflection of cloud in lake is
(h + 3) because in lake platform height is also added to reflection of cloud.
So,angle of depression is different  in the lake from the angle of elevation of the cloud above the surface of a lake

Alternate Method
False, we know that, if P is a point above the lake at a distance d, then the reflection of the point in the lake wouid be at the same distanced. Also, the angle of elevation and depression from the surface of the lake is same.
Here, the man is standing on a platform 3 m above the surface, so its angle of elevation to the cloud and angle of depression to the reflection of the cloud is not same.

Question 9:
The value of 2 sinθ can be a + \(\frac { 1 }{ a } \) where a is a positive number and a ≠ 1.
Solution:
False
Given, a is a positive  number  and a ≠ 1, then AM > GM

Which is not possible.
Hence,the value of 2 sinθ can not be a + \(\frac { 1 }{ a } \)

Question 10:
cosθ = \(\frac{{a}^{2}+{b}^{2}}{2ab}\) where a and b are two distinct numbers such that ab > 0.
Solution:
False
Given, a and b are two distinct numbers such that ab > 0.
Using,                                      AM > GM

Question 11:
The angle of elevation of the top of a tower is 30°. If the height of the tower is doubled, then the angle of elevation of its top will also be doubled.
Solution:
False

Question 12:
If the height of a tower and the distance of the point of observation from its foot, both.are increased by 10%, then the angle of elevation of its top remains unchanged.
Solution:
True

Hence,the required angle of elevation of its top remains unchanged

Exercise 8.3 Short Answer Type Questions

Question 1:

Solution:

Question 2:

Solution:

Question 3:
If tan A = \(\frac { 3 }{ 4 } \),then sin A cos A = \(\frac { 12 }{ 25 } \)
Solution:

Question 4:
(sinα + cosα) (tanα + cotβ) = secα + cosecβ
Solution:

Question 5:
(√3 + 1) (3 – cot 30°) = tan³ 60° – 2sin60°
Solution:

Question 6:

Solution:

Question 7:
tanθ + tan (90° – θ) = sec θ sec (90° – θ)
Solution:

Question 8:
Find the angle of elevation of the sun when the shadow of a pole h m high is √3 h m long.
Solution:
Let the angle of elevation of the sun is θ.
Given,        height of pole = h

Question 9:
If √3 tanθ = 1, then find the value of sin² θ – cos² θ
Solution:

Question 10:
A ladder 15 m long just reaches the  top of a vertical wall.If the ladders makes an angle of 60° with the wall,then find the height of the wall.
Solution:
Given that,the height of the ladder = 15 m
Let the height of the vertical wall = h
and the ladder makes an angle  of elevation 60° with the wall i.e θ = 60°

Question 11:
Simplify (1 + tan² θ) (1 – sinθ) (1 + sinθ)
Solution:

Question 12:
If 2 sin² θ – cos² θ = 2, then find the value of θ.
Solution:

Question 13:

Solution:

Question 14:
An observer 1.5 m tall is 20.5 m away from a tower 22 m high. Determine the angle of elevation of the top of the tower from the eye of the observer.
Solution:

which may be either positive or negative.Hence, required angle of elevation of the top of the tower from the eye of the observer is 45°.

Question 15:
Show that tan⁴ θ + tan² θ = sec⁴ θ – sec² θ.
Solution:
LHS = tan⁴ θ + tan² θ = tan² θ (tan² θ+1)
= tan² θ.sec² θ                                                                                 [∴ sec²θ = tan² θ+1]
= (sec² θ-1) . sec² θ                                                                         [∴tan²θ = sec² θ – 1]
= sec⁴ θ – sec² θ = RHS

Exercise 8.4 Long Answer Type Questions

Question 1:
If cosec θ+ cot θ = p, then prove that cosθ = \(\frac { { p }^{ 2 }-1 }{ { p }^{ 2 }+1 } \)
Solution:

Question 2:
Prove that \(\sqrt{{sec}^{2}\theta +{cosec}^{2}\theta}\)= tan θ + cot θ.
Solution:

Question 3:
The angle of elevation of the top of a tower from certain point is 30°. If the observer moves 20 m towards the tower, the angle of elevation of the top increases by 15°. Find the height of the tower.
Solution:

Question 4:
If 1 + sin² θ = 3 sin0 cos0, then prove that tan0 = 1 or \(\frac{1}{2}\)
Solution:

Question 5:
If sinθ + 2 cosθ = 1, then prove that 2 sinθ – cosθ = 2.
Solution:

Question 6:
The angle of elevation of the top of a tower from two points distant s and t from its foot are complementary. Prove that the height of the tower is √st.
Solution:
Let the height of the tower is h.
and                                  ∠ABC = θ
Given that,                             BC = s, PC =t
and angle of elevation on both positions are complementary.
i.e ,                                 ∠APC = 90°- θ

So, the required height of the tower is √st.
Hence proved.

Question 7:
The shadow of a tower standing on a level plane is found to be 50 m longer when Sun’s elevation is 30° than when it is 60°. Find the height of the tower.
Solution:
Let the height of the tower be h and RQ = x m


Hence, the required height of tower is 25√3 m.

Question 8:
A vertical tower stands on a horizontal plane and is surmounted by a vertical flag staff of height h. At a point on the plane, the angles of elevation of the bottom and the top of the flag staff are α and β respectively. Prove that the height of the tower is \(\left(\frac{h\quad tan\alpha}{tan\beta-tan\alpha}\right)\)
Solution:
Let the height of the tower be H and OR = x
Given that, height of flag staff = h = FP and ∠PRO = α, ∠FRO = β

Question 9:
If tan θ + sec θ = l, then prove that secθ = \(\frac { { l }^{ 2 }+1 }{ 2l } \).
Solution:

Question 10:
If sinθ + cosθ = p and sec θ + cosec θ = q, then prove that q(p²-1) = 2p.
Solution:

Question 11:
If a sinθ+ b cosθ = c, then prove that a cosθ – b sinθ = \(\sqrt {a^{2}+{b}^{2}-{c}^{2}}\).
Solution:

Question 12:

Solution:

Question 13:
The angle of elevation of the top of a tower 30 m high from the foot of another tower in the same plane is 60° and the angle of elevation of the top of the second tower from the foot of the first tower is 30°. Find the distance between the two towers and also the height of the tower.
Solution:
Let distance between the two towers = AB = x m
and height of the other tower = PA = h m
Given that, height of the tower = QB = 30 m and ∠QAB = 60°, ∠PBA = 30°

Hence, the required distance and height are 10√3 m and 10 m, respectively.

Question 14:
From the top of a tower h m high, angles of depression of two objects, which are in line with the foot of the tower are a and β (β > a). Find the distance between the two objects.
Solution:
Let the distance between two objects is x m,
and CD = y m.
Given that, ∠BAX = α = ∠ABD,              [alternate angle]
∠CAY = p = ∠ACD            [alternate angle]

Question 15:
A ladder against a vertical wall at an inclination a to the horizontal. Sts foot is pulled away from the wall through a distance p, so that its upper end slides a distance q down the wall and then the ladder makes an angle B to the horizontal. Show that \(\frac {p}{q}=\frac{cos\beta -cos\alpha}{sin\alpha-sin\beta}\)
Solution:

Question 16:
The angle of elevation of the top of a vertical tower from a point on the ground is 60° From another point 10 m vertically above the first, its angle of elevation is 45°. Find the height of the tower.
Solution:
Let the height the vertical tower, OT = H

Hence, the required height of the tower is 5 (√3 + 3) m,

Question 17:
A window of a house is h m above the ground. Form the window, the angles of elevation and depression of the top and the bottom of another house situated on the opposite side of the lane are found to be a and p, respectively. Prove that the height of the other house is h(1+tanα cot β)m.
Solution:

Question 18:
The lower window of a house is at a height of 2 m above the ground and its upper window is 4 m vertically above the lower window. At certain instant the angles of elevation of a balloon from these windows are observed to be 60° and 30°, respectively. Find the height of the balloon above the ground.
Solution:
Let the height of the balloon from above the ground is H.
A and OP=w₂R=w₁Q=x
Given that, height of lower window from above the ground = w₂P = 2 m = OR
Height of upper window from above the lower window = w₁w₂ = 4 m = QR

So, the required height is 8 m.
Hence, the required height of the ballon’from above the ground is 8 m.

NCERT Exemplar ProblemsNCERT Exemplar MathsNCERT Exemplar Science