plz solve this... its urgent ....
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Seg AB is a diameter
angle inscribed in a semicircle is 90°
m<ACB = 90°
In triangle ABC
m<A + m<B + m<C = 180 (sum of measure of all angles of triangle)
30+m<B + 90 = 180
120 m<B =180
m<B = 120-180 = 60°
<ABC = < PCA (angles on alternate seg)
therefore m<PCA = 60°
angle inscribed in a semicircle is 90°
m<ACB = 90°
In triangle ABC
m<A + m<B + m<C = 180 (sum of measure of all angles of triangle)
30+m<B + 90 = 180
120 m<B =180
m<B = 120-180 = 60°
<ABC = < PCA (angles on alternate seg)
therefore m<PCA = 60°
khushisingh35:
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