6.7.4 Angles and Tangents of Circles, PT3 Focus Practice - Mathematics

Question 9:

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

The value of x is

Solution:

∠OBC = 90^o

∠BOD = 2 × 50^o= 100^o

In quadrilateral BODC,

x^o= 360^o– ∠BOD – ∠OBC – 120^o

= 360^o– 100^o – 90^o – 120^o

= 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:

$\begin{array}{l} \cos 86^{o} = \frac{O M}{O V} \\ 0.070 = \frac{1}{O V} \\ O V = \frac{1}{0.070} \\ O V = 14.29 c m \\ ∴ P R = 14.29 - 5 - 4 \\ = 5.29 c m \end{array}$

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