Understanding Engineering Mathematics

to the opposite sideBC– see Figure 5.14. Since we already know that for two parallel
lines, alternate angles are equal, then we can saya=dandc=e.

AA

CBCB

a

b

ddee

ca

b

c

Figure 5.14Sum of the angles in a triangle.

Also, from angles on a straight line we have

a+b+c= 180 °

Hence
d+b+e= 180 °

which proves the result.
Another important property of triangles is thatan exterior angle of a triangle is equal
to the sum of the two opposite interior angles.
Again, the proof involves additional construction and is not difficult, as shown in
Figure 5.15.

ACD

B

acd e

b

Figure 5.15Exterior angle.

ACis ‘produced’ or extended toD. An additional line is also drawn throughCparallel to
AB. Then, by alternate angles we haved=b, and by corresponding angles we havee=a.
So,d+e=external angle atC=a+b, as required, i.e. BCD= BAC+ CBA.

Solution to review question 5.1.3

(i) Referring to Figure 5.3(i) we havea= 180 °− 70 °− 60 °= 50 ° by

angles in a triangle. Thenb= 70 °+ 60 °= 130 °by the external angle

result, or alternatively by angles on a straight lineb= 180 °−a=

180 °− 50 °= 130 °.

(ii) Since the triangle in Figure 5.3(ii) is isosceles,a= 65 °.Sob= 180 °−

2 × 65 °= 50 °andc=b+ 65 °= 115 °by the external angle result or

by angles on a linec= 180 °− 65 °= 115 °.