plz answer fast ITS VERY VERY URGENT
the tangent at a point C of a circle and diameter AB when extended intersects at P if angle PCA =110° find angle CBA
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17
Answer:
∠CBA = 70°
Step-by-step explanation:
From figure:
∠BCA = 90°
Given, ∠PCA = 110°.
⇒ ∠PCA = ∠PCB + ∠BCA
⇒ 110° = ∠PCB + 90°
⇒ ∠PCB = 20°
∴ ∠BAC = ∠PCB = 20°.
Now,
In ΔABC,
⇒ ∠CBA + ∠BCA + ∠CAB = 180°
⇒ ∠CBA + 90° + 20° = 180°
⇒ ∠CBA + 110° = 180°
⇒ ∠CBA = 70°.
Hope it helps!
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Join points C and O.
∠BCA = 90° [Since angle in a semi circle is 90°]
Also ∠PCO = 90° [Since radius ⊥ tangent]
From the figure we have,
∠PCA =∠PCO + ∠OCA
i.e. 110° = 90° + ∠OCA
Therefore, ∠OCA =20°
Now in ΔAOC, AO = OC [Radii]
So, ∠OCA = ∠OAC =20°
In ΔABC, we have
∠BCA = 90° & ∠CAB = 20°
Therefore, ∠CBA = 70°
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