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In the figure (ii) given below, O is the circumcenter of triangle ABC in which AC = BC. Given that ∠ACB = 56°, calculate

(i) ∠CAB

(ii) ∠OAC.

In the figure (ii) given below, O is the circumcenter of triangle ABC in which AC = BC. Given that ∠ACB = 56°, calculate ∠CAB, ∠OAC. Circles, ML Aggarwal Understanding Mathematics Solutions ICSE Class 10.

Circles

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Answer

(i) From figure,

In the figure (ii) given below, O is the circumcenter of triangle ABC in which AC = BC. Given that ∠ACB = 56°, calculate ∠CAB, ∠OAC. Circles, ML Aggarwal Understanding Mathematics Solutions ICSE Class 10.

AC = BC so,

∠CBA = ∠CAB (As angles of equal sides are equal)

In △ABC,

∠CAB + ∠CBA + ∠ACB = 180°
2∠CAB + 56° = 180°
2∠CAB = 180° - 56°
2∠CAB = 124°
∠CAB = 62°.

Hence, ∠CAB = 62°.

(ii) OC is the radius of the circle. OC bisects ∠ACB.

∠OCA = 12\dfrac{1}{2}∠ACB = 12×\dfrac{1}{2} \times 56° = 28°.

Now in △OCA,

OA = OC (Radius of the same circle)

∠OAC = ∠OCA = 28°.

Hence, ∠OAC = 28°.

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