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6.7.3 Angles and Tangents of Circles, PT3 Focus Practice



Question 7:


In figure above, ABCD is a tangent to the circle CEF at point C. EGC is a straight line. The value of y is
 
Solution:
CEF=DCF=70AEG+70+210=360AEG=80In cyclic quadrilateralABGE,ABG+AEG=180y+80=180y=100


Question 8:

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 = 360– ∠ABO  – ∠BOE – 35o
= 360– 90o – 80o – 35= 155o
yo+ ∠AEO = 180
yo+ 155o = 180o
y= 180o  – 155o
y o = 25

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