6.7.4 Angles and Tangents of Circles, PT3 Focus Practice


Question 9:
 

In figure above, ABC is a tangent to the circle centre O, at point B.
The value of is

Solution:
OBC = 90
BOD = 2 × 50o = 100o
In quadrilateral BODC,
x= 360– ∠BOD – ∠OBC – 120o
= 360– 100o – 90o – 120o
= 50


Question 10:


The figure above shows two circles with respective centres O and V. AB is a common tangent to the circles. OPRV is a straight line. The length, in cm, of PR is 

Solution:

cos 86 o = O M O V 0.070 = 1 O V O V = 1 0.070 O V = 14.29 c m P R = 14.29 5 4 = 5.29 c m

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