in the given figure , AC is the diameter of the circle with centre O. If <ADE =30° ;< DAC= 35° and <CAB = 40°.Find (i) <ACD
(ii) < ACB
(iii)<DAE
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Solution :
In triangle ACB,
<ACB = 90 (Angle in a semicircle)
Sum of opposite angles in a quadrilateral = 180
<ADC + <ABC = 180
120 + <ABC = 180
<ABC = 60
<ACB + <CAB + <ABC = 180
x + 90 + 60 = 180
x + 150 = 180
x = 180 - 150
x = 30
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Answered by
74
In triangle ACB,
<ACB = 90 (Angle in a semicircle)
Sum of opposite angles in a quadrilateral = 180
<ADC + <ABC = 180
120 + <ABC = 180
<ABC = 60
<ACB + <CAB + <ABC = 180
x + 90 + 60 = 180
x + 150 = 180
x = 180 - 150
x = 30
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