Question

in the given figure, AB is the diameter. Find
angle BAC​

Answer 1

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As AB is the diameter so ACB is a semicircle.

Os <ACB =90° {angle in semi circle is always90°}

As ABCD is a cyclic quadrilateral so

<D+<B=180°

<B+110°=180°

<B=180°-110°

<B= 70°

Now in triangle ABC,

<ABC+<BAC+<ACB=180° [Angle suk property]

<BAC + 70° + 90° = 180°

<BAC = 180° - 160°

<BAC = 20°

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Answer 2

Answer:

As AB is the diameter so ACB is a semicircle.

Os <ACB =90° {angle in semi circle is always90°}

As ABCD is a cyclic quadrilateral so

<D+<B=180°

<B+110°=180°

<B=180°-110°

<B= 70°

Now in triangle ABC,

<ABC+<BAC+<ACB=180° [Angle suk property]

<BAC + 70° + 90° = 180°

<BAC = 180° - 160°

<BAC = 20°