Question 7:
![](http://spmmathematics.blog.onlinetuition.com.my/wp-content/uploads/sites/5/2015/01/Picture17-1.png)
Solution:
![](http://spmmathematics.blog.onlinetuition.com.my/wp-content/uploads/sites/5/2015/01/Picture17-1.png)
In figure above, ABCD is a tangent to the circle CEF at point C. EGC is a straight line. The value of y is
Question 8:
![](http://spmmathematics.blog.onlinetuition.com.my/wp-content/uploads/sites/5/2015/01/Picture31.png)
![](http://spmmathematics.blog.onlinetuition.com.my/wp-content/uploads/sites/5/2015/01/Picture31.png)
In figure above, ABC is a tangent to the circle centre O at B. AED is a straight line.
Find the value of y.
Solution:
∠ABO = 90o
∠BOE = 2 × 40o = 80o
In quadrilateral AEOB,
∠AEO = 360o – ∠ABO – ∠BOE – 35o
= 360o – 90o – 80o – 35o = 155o
yo+ ∠AEO = 180o
yo+ 155o = 180o
yo = 180o – 155o
y o = 25