pls help me solve this Q
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Since QUADRILATERAL ABCD is cyclic, so
angle B = 180°–angle D
=› angles B = 50°
Now the ∆ABC is a triangle in a semi circle.
so, angle C = 90°
Then apply angle sum property of triangle in ∆ABC to get,
angle A =180°–(90°+50°)
Solve it. You can.
angle B = 180°–angle D
=› angles B = 50°
Now the ∆ABC is a triangle in a semi circle.
so, angle C = 90°
Then apply angle sum property of triangle in ∆ABC to get,
angle A =180°–(90°+50°)
Solve it. You can.
yashu114:
that it
we need to find angle CAB
It is angle CAB, I have just written A in short because I have mentioned "In ∆ABC".
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