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TRIGONOMETRY
This chapter comes under Paper-II. From this chapter 5 Marks question -1 (1×5 = 5M),
4 Marks questions - 2 (2×4 = 8M), 2 Marks questions - 2 (2×2 = 4M), 1 Mark question - 1
(1×1 = 1M) and 6 objective bits will be given. Altogether we can score 21 marks from it. This
chapter is not only very important for S.S.C exam but also to higher classes like Intermediate
and Engineering etc. The important material relevant to this topic is given below for the sake
of the students who are going to appear S.S.C public exams.
♦ Trigonometry is derived from three "Greek" roots ie Tri + Goma + Metron. Here Tri-means
♦
♦
♦
♦
♦
♦
three, Gomia means angle and Metron means measurement, this Trigonometry is the
study of "Three angle measurement".
Trigonometry is an analytical study of a three angled Geometric figure ie "triangle".
"Hipparchus" established the relationship between the sides and angles of a triangle.
Radian: The angle subtended by an arc of length equals to the radius of the circle at it's
centre is called 'radian'.
2πc = 360°, πc = 180°
90° = 100g = πc/2 or 180° = 200g = πc
The relationship among D, G, and C is
r
D
G
C
D
G
C
or
=
=
=
=
90 100 π / 2
180 200 π
1c
r
r
♦ If l is the length of arc of a sector, r is the radius, θ is the angle at the centre of sector
then the relationship among them is l = rθ or r =
♦
♦
♦
♦
l
l
or θ = ,
θ
r
Here θ should be radian only.
1° = 0.01746 radian ; 1c = 58°16'
3πc/4 = 135° (degrees)
120° = 400g/3 (grades)
270° = 3πc/2 (radians)
Problem: 1
(1 Mark)
A minute hand of a table clock is 3 cm long. How far it's tip move in 20 minutes?
Sol:
The length of minutes hand
r = 3 cm
Angle made by minutes hand in 60 minutes ---- 360°
20 minutes ------- ?(θ)
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6
θ=
20 × 360
πc
2π c
= 120° = 120 ×
=
180
3
60
1
r = 3cm
We know that l = rθ
θ
The distance travelled by tip of minutes hand
l = rθ
2π
= 3×
3
= 2×
=
r=?
22
7
44
cm
7
Problem: 2
(1 Mark)
A wheel makes 360 revolutions in 1 minute. Through how many radians does it turn in 1
second.
Sol:
A wheel makes
in 1 minute
→ 360 revolutions
i.e. 60 seconds → 360 × 2πc
6
1 second
→
360
× 2πc
60
1
1 second
→ 12πc
∴ The wheel makes 12πc in 1 second.
Trigonometric Ratio's:
♦ sin θ =
C
side opp.to θ BC
=
hypotenuse AC
♦ cos θ =
side adjacent to θ AB
=
hypotenuse
AC
♦ tan θ =
side opp.to θ BC
=
side adj.to θ AB
♦ cot θ =
side adj.to θ AB
=
side opp.to θ BC
♦ sec θ =
hyp
AC
=
sideadj.to θ AB
hyp.
Opp.side
θ
A
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adj.side
B
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♦ co sec θ =
♦ tan θ =
hyp
AC
=
sideopp.to θ BC
sin θ
cos θ
;cot θ =
cos θ
sin θ
♦ sinθ.cosθ = 1; cosθ.secθ = 1 ; tanθ.cotθ = 1
Trigonometric Values
θ
ratio
0°
30°
sin
0
1/2
1/√2
√3/2
1
cos
1
√3/2
1/√2
1/2
0
tan
0
1/√3
1
√3
∞
cot
∞
√3
1
1/√3
0
sec
1
2/√3
√2
2
∞
cosec
∞
2
√2
2/√3
45°
60°
Trigonometric Identities:
I. sin2θ + cos2θ = 1 ; sin2θ = 1 – cos2θ ; cos2θ = 1 – sin2θ
II. sec2θ – tan2θ = 1 ; sec2θ = 1 + tan2θ ; sec2 – 1 = tan2θ
III. cosec2θ – cot2θ = 1 ; cosec2θ = 1 + cot2θ ; cosec2θ – 1 = cot2θ
Problem: 3
(2 Marks)
If tan (A − B ) =
1
1
, sin A =
find B in circular measure.
3
2
Sol:
tan(A–B) = 1/√3
tan (A − B ) = tan 30°
A – B = 30° ____ (1)
sinA = 1/√2
sin A = sin 45°
A = 45° ____ (2 )
Substitute (2) in (1)
45° – B = 30°
–B = 30° – 45°
–B = –15°
B = 15°
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90°
1
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1
Now A = 45° = 45 ×
πc
180c
πc
=
4
4
1
πc
πc
=
B = 15° = 15 ×
180 12
12
Problem: 4
πc
πc
πc
2
3
Find the value of 32cot
− 8 sec
+ 8cot
4
3
6
2
Sol:
c
c
πc
2 π
3 π
− 8 sec
+ 8cot
32cot
4
3
6
2
= 32cot 2
45°
60°
30°
1
1
1
180°
180°
180°
− 8 sec 2
+ 8cot 3
4
3
6
= 32 cot2 45° – 8 sec2 60° + 8 cot3 30°
= 32(1)2 – 8(2)2 + 8.(√3)3
= 32 – 8(4) + 8 × 3√3
= 32 − 32 + 24 3
= 24√3
Problem: 5
If 4 cotA = 3 then find
sin A + cos A
sin A − cos A
C
Sol:
4 cot A = 3
sideadj.to θ
sideopp.to θ
By Pythagorean Theorem,
AC2 = AB2 + BC2
AC2 = 32 + 42
AC2 = 9 + 16
AC2 = 25 ⇒ AC = 5
5
4
⇒ cot A = 3 / 4 =
sin A =
θ
A
4
3
, cos A =
5
5
4 3
+
sin A + cos A 5 5
=
sin A − cos A 4 − 3
5 5
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3
B
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=
7/ 5
=7
1/ 5
Problem: 6
(2 Marks)
Show that
1 − cos θ
= cos ecθ − cot θ
1 + cos θ
Proof:
L.H.S
=
1 − cos θ
1 + cos θ
=
1 − cos θ 1 − cos θ
×
1 + cos θ 1 − cos θ
=
(1 − cos θ )2
1 − cos2 θ
(1 − cos θ )2
1 − cos2 θ
=
1 − cos θ
sin2 θ
=
1 − cos θ
sin θ
=
1
cos θ
−
sin θ sin θ
= cosecθ – cotθ
= R.H.S
Problem: 7
(4 Marks)
Prove that
tan θ + sec θ − 1 1 + sin θ
=
tan θ − sec θ + 1
cos θ
Proof:
tan θ + sec θ − 1
tan θ − sec θ + 1
=
tan θ + sec θ − sec 2 θ − tan2 θ 
tan θ − sec θ + 1
∵ sec 2 θ − tan2 θ = 1


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=
(tan θ + sec θ ) − (sec θ − tan θ )(sec θ + tan θ )
=
(tan θ + sec θ )1 − (sec θ − tan θ )
=
(tan θ + sec θ )(1 − sec θ + tan θ )
tan θ − sec θ + 1
tan θ − sec θ + 1
tan θ − sec θ + 1
=
tan θ + sec θ
1
=
sin θ
1
+
cos θ cos θ
=
sin θ + 1
cos θ
=
1 + sin θ
cos θ
Problem: 8
(4 Marks)
P2 − 1
If secθ + tanθ = P then show that sin θ = 2
P +1
Proof:
Given that P = secθ + tanθ
R.H.S=
P2 − 1 (sec θ + tan θ ) − 1
=
P2 + 1 (sec θ + tan θ )2 + 1
2
sec 2 θ + tan 2 θ + 2sec θ tan θ − 1
=
sec 2 θ + tan 2 θ + 2sec θ tan θ + 1
=
sec 2 θ − 1 + tan 2 θ + 2sec θ tan θ
sec 2 θ + tan 2 θ + 1 + 2sec θ tan θ
=
tan2 θ + tan 2 θ + 2sec θ tan θ
sec 2 θ + sec 2 θ + 2sec θ tan θ
2 tan2 θ + 2sec θ tan θ
=
2sec 2 θ + 2sec θ tan θ
=
(
)
2 sec θ (sec θ + tan θ )
2 tan θ tan θ + sec θ
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sin θ
cos θ
=
1/ cos θ
= sinθ
= L.H.S
Problem: 9
(4 Marks)
Eliminate θ from x = a sinθ – b cosθ, y = a cosθ + b sinθ
Sol:
x = a sinθ – b cosθ, y = a cosθ + b sinθ
Consider x = a sinθ – b cosθ
Squaring on both sides
x2 = (a sinθ – b cosθ)2
x2 = a2 sin2θ + b2 cos2θ – 2ab sinθcosθ ____ (1)
Consider y = a cosθ + b sinθ
Squaring on both sides
y2 = (a cosθ + b sinθ)2
y2 = a2 cos2θ + b2 sin2θ + 2ab cosθ sinθ ____ (2)
(1) + (2)
x 2 + y2 = a2 sin2 θ + b2 cos2 θ − 2ab sin θ cos θ
+ a2 cos2 θ + b2 sin2 θ + 2ab cos θ .sin θ
⇒ x2 + y2 = a2(sin2θ + cos2θ) + b2 (cos2θ + sin2θ)
⇒ x2 + y2 = a2(1) + b2(1)
⇒ x2 + y2 = a2 + b2
Problem: 10
(1 Mark)
If x = secθ + tanθ, y = secθ – tanθ then find the relationship between x and y.
Sol:
x = secθ + tanθ
y = secθ – tanθ
Consider xy = (secθ + tanθ) (secθ – tanθ)
= sec2θ – tan2θ = 1
♦ sin(90–θ) = cosθ,
tan(90–θ) = cotθ,
sec(90–θ) = cosecθ,
♦ sin(–θ) = –sinθ,
tan(–θ) = –tanθ,
sec(–θ) = secθ,
cos(90–θ) = sinθ,
cot(90–θ) = tanθ
cosec (90–θ) = secθ
cos(–θ) = cosθ
cot(–θ) = –cotθ
cosec(–θ) = –cosecθ
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Problem: 11
(4 Marks)
Show that sec(270–θ) = –cosecθ
Sol:
L.H.S
sec(270 – θ)
= sec[90 + 180 – θ)
Let 180 – θ= x
= sec[90 + x]
= sec[90–(–x)]
= cosec(–x)
= –cosec x
= –cosec(180–θ)
= –cosec[90 + 90 – θ]
Let 90 – θ=y
= –cosec [90 + y]
= –cosec [90–(–y)]
= –sec(–y)
= –secy
= –sec(90–θ)
= –cosecθ
= R.H.S
♦
♦
♦
♦
♦
♦
sin(180 + θ) = –sinθ
cos(270 + θ) = –sinθ
tan(90 + θ) = –cotθ
sec(270 – θ) = –cosecθ
cot(360 + θ) = cotθ
cosec (360 – θ) = –cosecθ
Angle of Elevation:
If the object is at a heigher level than the eye then
the angle between the horizontal line drawn
through the observers eye and the line joining the
eye to an object is called "Angle of Elevation".
t
igh
e
lin
s
of
Angle of elevation
Eye
Horizontal line
Angle of Depression:
If the object is at a lower level than the eye, then the
angle between the horizontal line drawn through the
observers eye and the line joining the eye to any
object is called "Angle of Depression".
Angle of
depression
ht
ig
fs
eo
lin
object
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Horizontal line
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Problem: 12
(2 Marks)
Find the height of mountain cliff, if the angle of elevation of it's top, from a point 300m from
it's foot is found to be 60°.
Sol:
Let AB is the mountain cliff. AC is the distance between cliff and observing point.
B
∠BCA = 60°
In ∆ABC, tan60 ° =
AB
AC
AB
300
⇒ AB = 300√3 m
⇒ AB = 300 × 1.732
⇒ AB = 519.6m
∴ Height of mountain cliff = 519.6m
3=
Mountain Cliff
θ = 60°
A
300 m
C
Problem: 13
(4 Marks)
The upper part of a tree, broken by wind into two parts, makes an angle of 30° with the
ground. The top of the tree touches the ground at a distance of 20m from the foot of the tree.
Find the height of the tree before it was broken.
Sol:
Let AB is the tree before it was broken.
C is broken point.
CB' is broken part.
B
∠CB'A = 30°
y
AB' = 20m
Let AC = x, BC = y = B'C
AC
C
In ∆CB'A, tan30 ° =
AB '
1
x
=
3 20
30°
20
3
⇒x=
×
3
3
⇒x=
B'
20 3
m ____ (1)
3
In ∆CB'A, cos30 ° =
x
y
AB '
CB '
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20 m
A
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3 20
=
2
y
y=
20 × 2
3
y=
40 3
3× 3
40 3
____ (2 )
3
The height of tree before it was broken = x + y
y=
=
20 3 40 3
+
3
3
20
=
60 3
3
= 20√3
= 20 × 1.732
= 34.64m
Problem: 14
There are two temples, one on each bank of a river, just opposite to each other. One of the
temple A is 40m high. As observed from the top of this temple 'A', the angles of depression
of the top and foot of the other temple 'B' are 12°30' and 21°48' respectively. Find the width
of the river and height of the temple "B".
Sol:
x
P
12°30'
40 - h
21°48'
12°30'
Q
R
d
40 m
h
h
21°48'
A
d
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B
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AP & BQ are two temples.
AP = 40m
AB is the width of river.
∠QPX = 12°30' ⇒ ∠RQP = 12°30',
∠BPX = 21°48' ⇒ ∠ABP = 21°48'
Let AB = d ⇒ RQ = d
Let BQ = h ⇒ AR = h ⇒ PR = 40–h
40 − h
Consider ∆PQR, tan12°30' =
d
0.2217 =
40 − h
____ (1)
d
In ∆PBA, tan 21°48' = 40/d
0.4000 = 40/d
4000
d×
= 40
10000
1
4 0× 10, 000
d=
4 000
1
d = 100m
Substitute 'd' value in (1)
40 − h
0.2217 =
100
2217
× 100 = 40 − h
10, 000
2217
= 40 − h
100
22.17 = 40–h
h = 40 – 22.17
h = 17.83m
∴ Width of the river (d) = 100m.
The height of temple B = 17.83m
Objective Bits:
1. If sinθ = cosθ then θ = _______
2. If tanθ – secθ = 1 then tanθ + secθ = _______
3. cosec2 9°– cot2 9° = _______
4. sin 420° = _______
5. sin29° + sin281° = _______
6. If tanθ = a/b then sinθ = _______
7. If sin2θ = cos3θ then (i) tan 5θ =
(ii) cot 5θ =
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sin18 °
= _______
cos72 °
9. sin15° cos15° =
10. 1° = ______ radians
11. x = secθ – tanθ, y = secθ + tanθ then the relationship between x and y is ____
12. cos 240° = _____
13. The relationship among D, G, and C is
14. sin258° + cos258° =
15. An angle made by a side of a 24 sided regular polygon at it's centre is ____
8.
Answers:
πc
1. θ =
or 45°
4
2.
3.
4.
5.
6.
–1
1
√3/2
1
a
a 2 + b2
7. i) ∞
ii) 0
8. 1
9. 1/4
10. 0.01746
11. xy = 1
12. –1/2
D
G
C
D
G
C
or
=
=
=
=
13.
90 100 π / 2
180 200 π
14. 1
15. 15°
Assignment:
(2 Marks)
1. The angles of a triangle are in A.P and the greatest is three times the least. Then find
the angles of triangle in circular measure.
2. Prove that 4(sin430° + cos460°) –3 (cos245° – sin290°) = 2
(1
3.
4.
(4
Mark)
Find the value of cos0° + sin90° + √2.sin45°
Show that tan2θ + tan4θ = sec4θ – sec2θ
Marks)
cos θ
cos θ
+
=4
5. Solve
1 − sin θ 1 + sin θ
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(2 Marks)
6. Show that cos(270 + θ) = sinθ
(4 Marks)
7. A ladder 20m long is placed against a vertical wall of height 10m. Find the distance
between the foot of the ladder and the wall, also find the inclination of the ladder with the
horizontal.
(4 Marks)
8. An aeroplane at an altitude of 2500 mts observes the angles of depression of opposite
points on the two banks of a river to be 41°20' and 52°10'. Find the width of the river in
metres.
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