Basic Engineering Mathematics, Fifth Edition

Angles and triangles 173

62 

15 

A

B D

E

C 



Figure 20.23

In triangleABC,∠A+∠B+∠C= 180 ◦(the angles in
a triangle add up to 180◦).
Hence,∠C= 180 ◦− 90 ◦− 62 ◦= 28 ◦. Thus,
∠DCE= 28 ◦(vertically opposite angles).
θ=∠DCE+∠DEC(the exterior angle of a triangle is
equal to the sum of the two opposite interior angles).
Hence,∠θ= 28 ◦+ 15 ◦= 43 ◦.
∠αand∠DECare supplementary; thus,
α= 180 ◦− 15 ◦= 165 ◦.

Problem 22. ABCis an isosceles triangle in

which the unequal angleBACis 56◦.ABis extended

toDas shown in Figure 20.24. Find, for the

triangle,∠ABCand∠ACB. Also, calculate∠DBC

568

A

B C

D

Figure 20.24

Since triangleABCis isosceles, two sides – i.e.ABand
AC– are equal and two angles – i.e.∠ABCand∠ACB–
are equal.
The sum of the three angles of a triangle is equal to 180◦.
Hence,∠ABC+∠ACB= 180 ◦− 56 ◦= 124 ◦.
Since∠ABC=∠ACBthen

∠ABC=∠ACB=

124 ◦

2

= 62 ◦.

An angle of 180◦lies on a straight line; hence,
∠ABC+∠DBC= 180 ◦from which,
∠DBC= 180 ◦−∠ABC= 180 ◦− 62 ◦= 118 ◦.
Alternatively, ∠DBC=∠A+∠C (exterior angle
equals sum of two interior opposite angles),
i.e.∠DBC= 56 ◦+ 62 ◦= 118 ◦.

Problem 23. Find anglesa,b,c,dandein

Figure 20.25

558

628

e b

d c

a

Figure 20.25

a= 62 ◦andc= 55 ◦(alternate angles between parallel

lines).

55 ◦+b+ 62 ◦= 180 ◦(angles in a triangle add up to

180 ◦); hence,b= 180 ◦− 55 ◦− 62 ◦= 63 ◦.

b=d= 63 ◦(alternate angles between parallel lines).

e+ 55 ◦+ 63 ◦= 180 ◦(angles in a triangle add up to

180 ◦); hence,e= 180 ◦− 55 ◦− 63 ◦= 62 ◦.

Check: e=a= 62 ◦ (corresponding angles between

parallel lines).

Now try the following Practice Exercise

PracticeExercise 78 Triangles (answerson

page 348)

1. Namethetypesoftriangleshownindiagrams
   (a) to (f) in Figure 20.26.

(a) (b)

(e) (f)

(c)

81  15 

48 

66 

45 

45 

(d)

97 

39 

60 

5 5

53 

37 

Figure 20.26