Angle between tangent and radius is 90 degree.
QM is tangent to ʘP at point M and QN is tangent to ʘP at point.
Solve for the variable and determine the angle measures.
Problem 1:
Solution :
PMNQ is quadrilateral. Sum of interior angles of quadrilateral is 360.
Since NQ and MQ are tangents, m∠QNP = m∠QMP = 90
x + 1 + 2x - 7 + 90 + 90 = 360˚
3x - 6 + 180 = 360˚
3x = 360 - 174
3x = 186
x = 186/3
x = 62˚
m∠NQM = x + 1 = 62 + 1 m∠NQM = 63˚ |
m∠NPM = 2x - 7 = 2(62) - 7 = 124 - 7 m∠NPM = 117˚ |
Problem 2 :
Solution :
2x - 2 + 6x - 10 + 90 + 90 = 360˚
8x - 12 + 180 = 360˚
8x = 360 - 168
8x = 192
x = 192/8
x = 24˚
m∠MQN = 2x - 2 = 48 - 2 m∠MQN = 46˚ |
m∠NPM = 6x - 10 = 6(24) - 10 = 144 - 10 m∠NPM = 134˚ |
Problem 3 :
Solution :
117˚ + 52˚ + a = 180˚ 169˚ + a = 180˚ a = 180˚ - 169˚ a = 11˚ |
a + b = 90˚ 11 + b = 90˚ b = 90˚ - 11˚ b = 79˚ |
In the triangle,
b + b + d = 180
2b + d = 180
2(79) + d = 180
158 + d = 180
d = 180 - 158
d = 22
Using alternate segment theorem,
b = c = 79
Problem 4 :
In the diagram, TA and TB are tangents. Find the angles a, b and c.
Solution :
Angles at the circumference of a circle subtended by the same arc are equal.
b = 36˚
74˚ + 74˚ + c = 180˚
148 + c = 180˚
c = 180˚ - 148˚
c = 32˚
Using alternate segment theorem,
a = 74
Problem 5 :
AT and BT are tangents to the circle, centre
C.
P is a point on the circumference, as shown.
∠BAT = 65˚
Calculate the size of
a)x b)y c)z
Solution :
AT = BT
∠ABT = ∠BAT = 65˚
∠ABT + ∠BAT + ∠ATB = 180˚
65 + 65 + z = 180˚
z = 180˚ - 130˚
z = 50˚
∠CAB + ∠CBA + ∠ACB = 180˚
25˚ + 25˚ + y = 180˚
50 + y = 180˚
y = 180 - 50
y = 130˚
∠APB = 1/2∠ACB
= 1/2(130˚)
∠APB = 65˚
x = 65˚
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