Need help on this math problem of parallelograms.
1 answer:
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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
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Solving this equation
<h2>-7.4 + E = 9.11</h2><h2>E = 16.51</h2>