An interior angle of a triangle is formed by two sides of the triangle. An exterior angle is formed by one side of the triangle and the extension of an adjacent side.
Each exterior angle has two remote interior angles. A remote interior angle is an interior angle that is not adjacent to the exterior angle.
Extend the base of the triangle and label the exterior angle as ∠4.
The Triangle Sum Theorem states :
m∠1 + m∠2 + m∠3 = 180 -----(1)
So, m∠3 + m∠4 = 180 -----(2)
(2) = (1)
m∠1 + m∠2 + m∠3 = m∠3 + m∠4
m∠1 + m∠2 = m∠4
Problem 1 :
Find the value of x.
Solution :
Using exterior angle theorem :
2x - 8 = x + 31
Subtracting x on both sides.
2x - x - 8 = 31
Add 8 on both sides.
x = 31 + 8
x = 39
Problem 2 :
Solution :
Using exterior angle theorem :
7x + 1 + 38 = 10x + 9
7x + 39 = 10x + 9
Subtracting 39 on both sides.
7x - 10x + 39 = 9
Subtracting 39 on both sides.
-3x = 9 - 39
-3x = - 30
x = 10
Problem 3 :
Solution :
Since the given triangle is isosceles triangle,
x = y
x + y = 130
x + x = 130
2x = 130
x = 130/2
x = 65
Problem 4 :
Solution :
Since the given triangle is isosceles triangle,
∠y = 35
x + y = 180
x + 35 = 180
x = 180 -35
x = 145
Problem 5 :
Find m∠ACB, m∠BCD and m∠DCE.
Solution :
In triangle ABC,
∠A + ∠B + ∠ACB = 180
78 + 58 + ∠ACB = 180
136 + ∠ACB = 180
∠ACB = 180 - 136
∠ACB = 44
In triangle DCE,
∠D + ∠E + ∠ECD = 180
60 + 85 + ∠ECD = 180
∠ECD = 180 - 145
∠ECD = 35
∠ACB + ∠BCD + ∠DCE = 180
44 + 35 + ∠BCD = 180
∠BCD = 180 - 79
∠BCD = 101
Problem 6 :
Find m∠K, m∠L, m∠KML, m∠LMN.
Solution :
In triangle LKM :
∠K + ∠L + ∠KML = 180
2x + 3x + x = 180
6x = 180
x = 180/6
x = 30
∠L = 3x ==> 3(30) ==> 90
∠K = 2x ==> 2(30) ==> 60
∠KML = x ==> 30
∠LMN = 180 - 30 ==> 150
May 21, 24 08:51 PM
May 21, 24 08:51 AM
May 20, 24 10:45 PM