Question

O is the centre of the circle, EF is a tangent, angle BCE = 28°, angle ACD = 31°

Answer 1

Answer

a. 28˚

b. 76˚

c. 104˚

d. 56˚

Step-by-step explanation

Given,

∠BCE=28°  ∠ACD=31°  &  line AB=AC .

According To the Question,

• a. the angle between a chord and a tangent through one of the end points of the chord is equal to the angle in the alternate segment.(Alternate Segment Theorem) Thus, ∠BAC=28°

• b. We Know The Sum Of All Angles in a triangle is 180˚, 180°-∠CAB(28°)=152° and ΔABC is an isosceles triangle, So 152°/2=76˚

        thus , ∠ABC=76° .

• c. We know the Sum of all angles in a triangle is 180° and opposite angles in a cyclic quadrilateral(ABCD) add up to 180˚,

Thus, ∠ACD + ∠ACB = 31° + 76° ⇔ 107°

Now, ∠DCB + ∠DAB = 180°(Cyclic Quadrilateral opposite angle)

∠DAB = 180° - 107° ⇔ 73°

& We Know, ∠DAC+∠CAB=∠DAB ⇔ ∠DAC = 73° - 28° ⇔ 45°

Now, In Triangle ADC Sum of angles in a triangle is 180°

∠ADC = 180° - (31° + 45°)  ⇔  104˚

   

• d. ∠COB = 28°×2 ⇔ 56˚ , because With the Same Arc(CB) The Angle at circumference are half of the angle at the centre  

For Diagram, Please Find in Attachment