Question 9:
![](http://spmmathematics.blog.onlinetuition.com.my/wp-content/uploads/sites/5/2015/01/Picture32.png)
In figure above, ABC is a tangent to the circle centre O, at point B.
The value of x is
Solution:
∠OBC = 90o
∠BOD = 2 × 50o = 100o
In quadrilateral BODC,
xo = 360o – ∠BOD – ∠OBC – 120o
= 360o – 100o – 90o – 120o
= 50
Question 10:
Solution:
![](http://spmmathematics.blog.onlinetuition.com.my/wp-content/uploads/sites/5/2015/01/Picture29.png)
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
![](http://spmmathematics.blog.onlinetuition.com.my/wp-content/uploads/sites/5/2015/01/Picture30.png)