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15
Angles in a Triangle
By studying this chapter you will be able to achive the following competencies.
« Proving the geometrical theorem of the exterior angle formed by
producing a side of a triangle is equal to the sum of the two interior
opposite angles and solving problems using it.
« Solving problems related to interior and exterior angles of a triangle.
0
« Proving that the sum of the interior angles of a triangle is equal to 180 .
15.1 Interior angles and exterior angles of a triangle.
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You have learnt earlier how to name the three angles of
the triangle ABC.
They are, ABC, BCA, CAB. These are the interior
angles of triangle ABC.
When the side BC is produced to D, the exterior angle
at the vertex C is ACD . When the side CA is produced
to E, the exterior angle at A is BAE.
C
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B
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E
A
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When ACD is considered as the exterior angle
then angles CAB and ABC are the interior
opposite angles to the exterior angle ACD .
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Similarly, CBA and BCA are the interior opposite
angles to the exterior angle BAE .
B
C
D
P
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Q
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In triangle PQR, the side PQ is produced to S and
angles RPQ and PRQ are the interior opposite
angles to the exterior angle SQR .
R
S
Theorem
The exterior angle formed by producing a side of a triangle is equal to the
sum of the two interior opposite angles.
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Activity 2
In triangle XYZ, the side YZ is produced to T. Draw ZP
parallel to YX through Z.
P
Considering the parallel lines XY and PZ and the
transversal XZ,
X
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T
name an alternate angle equals to XZP and
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Z
a corresponding angle equal to PZT.
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Hence name a pair of angles equal to the sum of ,
XZP + PZT, i.e., to XZT
Hence check whether the above theorem is true.
Y
Let us prove the above theorem systematically.
The proof of the above theorem.
A
B
X
D
C
:- ACD = CAB + ABC
Construction
:- CX is drawn parallel to BA through C.
Proof;
:- ACX = CAB —— (1) ^BA // CX, and alternate angles)
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To prove that
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:- The side BC of the triangle ABC is produced to D.
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Data
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XCD = ABC —— (2) ^BA // CX, and corresponding angles&
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(1) + (2) ACX + XCD = CAB + ABC
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ACX + XCD = ACD (ACX and XCD are a pair of adjacent angles)
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\ ACD = CAB + ABC
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Hence in triangle ABC, the exterior angle ACD formed by producing side BC to
D is equal to the sum of the interior opposite angles CAB and ABC.
Example 1
Find the magnitude of p in the given diagram.
P
p + 75o = 130o
o
o
p = 130 - 75
o
p = 55
75°
130°
Example 2
Find the magnitude of a in the given diagram and thereby find the value of the
exterior angle shown in the diagram
o
o
o
a + 10 + 58
= 3a + 32
o
o
a + 68 = 3a + 32
3a +32°
o
o
68 - 32 = 3a - a
o
\ 2a = 36
a = 36
°
0
58°
2o
a +1
a = 18
Exterior angle
= 3a + 32
o
o
= 3 ´ 18 + 32
= 54o + 32o
o
= 86o
Exercise 15.1
1. Find the value of x in each diagram.
2x
27°
x°
3x
66°
108°
x°
63°
x°
72°
x°
69°
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(i)
(ii)
(iii)
(iv)
2. In D PQR, the side QR is produced to S, TP is drawn parallel to SQ through P,
Prove that TPR = 180° - (QPR + PQR)
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3.
r
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In the given diagram, if q = 2p, show that p = r
3
q
p
P
4. In the diagram if PQ // SR prove that
y=x+z
x
Q
y
z
S
R
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5. In D ABC, the side BC is produced to K. The bisectors of the interior angles
BAC and ABC intersect at O. AO is produced to meet BC at P. Prove that
ACK= 2BOP
15.2 Angles in a triangle
Theorem
The sum of the three interior angles of a triangle is two right angles.
Activity 2
Draw any triangle ABC. Separate the vertices A, B,
C as shown in the diagram. Paste the separated
pieces on a paper such that the vertices of the
triangle are placed at one point and the sides are
adjacent as shown in the diagram.
Observe whether POQ is a straight line.
What can you conclude? Let us prove
this theorem, systematically.
P
For Free Distribution
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b a c
O
Q
P
Proof of the theorem.
T
R
Q
S
:- PQR + QRP + RPQ = 180°
Construction
:- QR is produced to S, RT is drawn parallel to QP through R
Proof
:- RPQ = PRT —— (1) ^QP // RT, alternate angles&
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To prove that
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:- PQR is any triangle.
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Data
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RPQ + PQR = PRT + TRS
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(1) + (2)
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PQR = TRS —— (2) ^QP // RT, corresponding angles&
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By adding QRP to both sides of the equation,
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PQR + QRP + RPQ = QRP + PRT + TRS.
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As QRP +PRT + TRS = 180° (Supplementary adjacent
angles on the line QRS),
PQR + QRP + RPQ = 180°
According to the above proof, the sum of the three interior angles of a triangle
is two right angles.
Example 3
The magnitude of 3 angles of a triangle are in the ratio 2: 5: 11. Calculate the
magnitude of the largest angle and hence find what type of a triangle this is.
If the smallest angle is 2a, the medium angle is 5a and the largest angle is 11a.
Consider the sum of the angles
2a + 5a + 11a
= 180o
o
18a
= 180
a
a
The largest angle
= 180°
18
o
= 10
= 11a
= 11 ´ 10o As the largest angle is 1100, this is an
obtuse angled triangle.
= 110o
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Example 4
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A
(i) Express the exterior angle ACD of D ABC in terms of p.
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p + 20°
ACD = ABC + BAC (Exterior angle is equal to the sum
p + 35°
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of the interior opposite angles)
o
ACD = p + 35 + p + 20
B
o
C
D
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= 2p + 55o
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= 180o (Supplementary adjacent angles)
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= 180 - ACD
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ACB + ACD
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(ii) Write the magnitude of ACB in terms of p.
= 180o - (2p + 55o )
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\ ACB
ACB
= 180o - 2p - 55o
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= 125 - 2p
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(iii) If ACB = 65° , what is the value of p?
ACB = 65
o
65o
= 125o - 2p
2p
= 125 - 65
2p
= 60
60°
=
2
o
= 30
o
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p
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p
o
BAC
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= p + 20o
o
ABC
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= 30 + 20
BAC
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(iv) Find the magnitude of BAC and ABC
o
ABC
= 50o
= p + 35o
o
= 30 + 35
o
= 65o
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Example 5
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The side QR of D PQR is produced to S. The bisector of angle QPR intersects QR
at T. Prove that PQS + PRS = 2PTS .
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P
Side QR of triangle PQR is
produced to S. The bisector of
angle QPR intersects QR at T.
Q
PQS + PRS = 2PTS
T
R
PTS = PQS + QPT,
S
(Exterior angle is equal to the sum of the interior
opposite angles)
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To prove that :
Proof
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Data
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PQS = PTS - QPT
(1) (QPT = 1 QPR)
PQS = PTS - 1 QPR
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2
PRS = PTS + TPR (Exterior angle is equal to the sum of
interior opposite angles')
(2) (TPR = 1 QPR)
PRS = PTS +1 QPR
2
2
(1) + (2)
PQS + PRS = 2PTS
Example 6
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"In any quadrilateral, the sum of a pair of opposite angles is equal to the sum
of the exterior angles at the other two vertices."
Data :- In the quadrilateral ABCD, the side AB is produced to P and the side
CD is produced to Q
Q
D
To prove that :- BAD + BCD = CBP + ADQ
Construction :- Draw the diagonal AC
Proof :-
C
A
InD ABC,
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B
P
BAC + ACB = CBP - 1 (Sum of interior opposite
angles is equal to the exterior angle)
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In D ACD,
CAD + ACD = ADQ
2 (Above theorem)
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1 + 2 ,
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BAC + CAD + ACB + ACD = CBP + ADQ
BAD + BCD = CBP + ADQ
Using the theorems you have learnt in this lesson, write the following proofs clearly.
Exercise 15.2
1. In the diagrams given below, find the value of a
28°
a°
79°
48°
(i)
5a
a°
29°
2a
a°
105°
5a
(ii)
2a
(iii)
(iv)
2. The three interior angles of a triangle are in the ratio 3: 5: 7. Find the magnitude
of the smallest angle and the largest angle of the triangle. Will this be an acute
angled triangle?
3. In a right angled triangle, the two acute angles are in the ratio 3:7. Find the
magnitude of the smallest angle.
4. The magnitude of one angle of a triangle is 720 . The other two angles are in the
ratio 8:1. Find the magnitude of the largest angle. Show that this is an
obtuse angled triangle.
5. In a triangle the smallest angle and the largest angle are in the ratio 1:2, while
the largest angle and the medium angle are in the ratio 4:3. Calculate the
magnitude of the three angles of the triangle.
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6. In D PQR, if P + R = 128° and Q +R = 105° , Find the magnitude of the three
angles separately. Show that, PQR is an acute angled triangle.
8.
In the diagram, PQ//SR. The bisectors of
the angles PQR and QRS intersect each
other at O. Prove that QOR = 90°.
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If the diagonal AC of the quadrilateral ABCD bisects the angles DAB and ,
BCD, Prove that ABC = ADC .
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7.
D
H
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9. In the diagram, if , DEF = GDF Prove that
EDF = DGF .
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By using the above result or otherwise prove that
FDH = DGE.
G
F
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E
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10. In D XYZ, angle XYZ = 90°. The bisectors of the interior angles X and Z
intersect at P. Prove that XPZ = 135° '
11. By drawing one diagonal, prove that in any quadrilateral, the sum of the
interior angles is 3600.
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