Question 4:
Solution:
∠CEF=∠DCF=70∘∠AEG+70∘+210∘=360∘∠AEG=80∘In cyclic quadrilateralABGE,∠ABG+∠AEG=180∘y∘+80∘=180∘y=100
In figure above, ABCD is a tangent to the circle CEF at point C. EGC is a straight line. The value of y is
∠CEF=∠DCF=70∘∠AEG+70∘+210∘=360∘∠AEG=80∘In cyclic quadrilateralABGE,∠ABG+∠AEG=180∘y∘+80∘=180∘y=100
Question 5:
In figure above, PAQ is a tangent to the circle at point A. AEC and BED are straight lines. The value of y is
Solution:
∠ABD = ∠ACD = 40o
∠ACB = ∠PAB = 60o
y= 180o – ∠ACB – ∠CBD – ∠ABD
y= 180o – 60o – 25o– 40o = 55o
Question 6:
Solution:
In figure above, KPL is a tangent to the circle PQRS at point P. The value of x is
Solution:
∠PQS = ∠SPL= 55o
∠SPQ = 180o – 30o – 55o= 95o
In cyclic quadrilateral,
∠SPQ + ∠SRQ = 180o
95o+ xo = 180o
x = 85o