Problem 1 :
Solution :
∠ABE = 120°
∠ABD + ∠DBE = 120° ----(1)
∠CBD = 135°
∠CBE + ∠EBD = 135°
∠CBE + ∠DBE = 135° ----(2)
(2) - (1) :
(∠CBE + ∠DBE) - (∠ABD + ∠DBE) = 135° - 120°
∠CBE - ∠ABD = 15°
∠CBE = 15° + ∠ABD
From the figure, we know that
∠ABE + ∠EBC = 180°
120° + 15° + ∠ABD = 180°
∠ABD = 180° - 135°
∠ABD = 45°
By applying the value of ∠ABD in (1), we get
45 + ∠DBE = 120°
∠DBE = 120° - 45°
∠DBE = 75°
Problem 2 :
Solution :
Since r || t, 5y - 9 and 3x are alternate interior angle.
5y - 9 = 3x
3x - 5y = -9 ----(1)
5y - 9 and 5x + 4 are linear pair.
5x + 4 + 5y - 9 = 180
5x + 5y - 5 = 180
5x + 5y = 185
x + y = 37 ----(2)
(1) + 5(2) :
3x - 5y + 5(x + y) = -9 + 185
3x - 5y + 5x + 5y = 176
8x = 176
x = 176/8
x = 22
Problem 3 :
Solution :
Since n and m are parallel, alternate interior angles are equal.
Since l and m are parallel, alternate interior angles are equal.
x + y + 70 = 360
x + y = 360 - 70
x + y = 290
Problem 4 :
Solution :
In a triangle, sum of interior angles of triangle is 180°.
a + c + 180 - b = 180
50 + c + 180 - 120 = 180
230 - 120 + c = 180
110 + c = 180
c = 180 - 110
c = 70
Problem 5 :
Solution :
Since l and m are parallel, alternate interior angles are equal.
In a triangle, sum of interior angles of triangle is 180°.
55 + 90 + x = 180
145 + x = 180
x = 180 - 145
x = 35
Problem 6 :
Solution :
Since lines l and m are parallel, corresponding angles will be equal.
In the line l,
75 + 50 + x = 180
125 + x = 180
x = 180 - 125
x = 55
Problem 7 :
Solution :
108° + ∠BAD = 180°
∠BAD = 180° - 108°
∠BAD = 72°
∠ABC = 180° - 72°
∠ABC = 108°
∠ABC = ∠ABD + ∠DBC
Since ∠ABD = ∠DBC,
∠ABC = ∠ABD + ∠ABD
∠ABC = 2∠ABD
108° = 2∠ABD
Divide both sides by 2.
54° = ∠ABD
In triangle ABD,
72° + 54° + x° = 180°
x + 126 = 180
x = 54
Problem 8 :
Solution :
x + y = 180 ----(1)
Corresponding angles are equal,
2x + 15 = y
2x-y = -15 ----(2)
(1) + (2) :
x + 2x = 180 - 15
3x = 165
x = 55
Substitute x = 55 in (1).
55 + y = 180
y = 180 - 55
y = 125
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