Question Is From Circle Chapter Of Class 10th.
Please Solve This.
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please join DE and refer your diagram
consider ΔCED
∠DEB=90° (angles in a semicircle)
CE=DE
∴∠EDC=∠ECD =45° (sum of angles in a triangle(90+2x))
and ∠DCG=90° (tangent makes 90° with the diameter)
∴∠BCG=75°
now find the angles
∠BCE=45°-15°=30°
∠GCE=75°-30°=45°
all the best
consider ΔCED
∠DEB=90° (angles in a semicircle)
CE=DE
∴∠EDC=∠ECD =45° (sum of angles in a triangle(90+2x))
and ∠DCG=90° (tangent makes 90° with the diameter)
∴∠BCG=75°
now find the angles
∠BCE=45°-15°=30°
∠GCE=75°-30°=45°
all the best
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