Foundation
Mathematics Foundation
Tangent to Circle
Question
In the figure given below ∠AOB = 80º, arc AC = arc BC, then ∠CAB is equal to
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Solution
We have,
∠ACB = ∠AOB
∠ACB = (80º)
= 40º
Now in triangle ABC
∴ AC = BC
∴ ∠CAB = ∠CBA --------(1)
and ÐACB + ∠CBA + ∠CAB = 180
∠CBA + ∠CAB = 180º – 40º
∠CBA + ∠CAB = 140º --------(2)
from equation (1) and (2)
2 ∠CAB = 140º
∠CAB = 70º
