Question

10 In the given figure, FG is a tangent to the circle with centre A. If ZDCB = 15° and CE = DE, then find ZGCE and ZBCE. ​

Answer 1

Answer:

45deg and 30deg

Step-by-step explanation:

plz check attachment for solution

Answer 2

Concept Introduction:

Radius of a circle is perpendicular to the tangent meeting at one point.

triangle in a semicircle is right angled triangle.

Given:

∠DCB = $15$°, CE = DE

To Find:

We have to find the value of, ∠GCE and ∠BCE. ​

Solution:

According to the problem,

ΔCDE is triangle in semicircle hence it is right angle triangle where  ∠CED = $90$° And it is given that CE = DE so it becomes isosceles right angled triangle ∴∠ECD = ∠EDC = $45$°

It is given that ∠DCB = $15$° and ∠DCG = $90$°(∵Radius of a circle is perpendicular to the tangent meeting at one point.)

⇒∠GCE + ∠ECD = $90$°

⇒∠GCE +  $45$° = $90$°

⇒∠GCE = $45$°

Now ∠ECD = ∠BCE + ∠DCB

⇒  $45$° = ∠BCE + $15$°

⇒∠BCE = $30$°

Final Answer:

The value of ∠BCE = $30$° and ∠GCE = $45$°

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