AB is a chord of the circle and AOC is its diameter such that angle ACB = 50°. If AT is the tangent to the circle at the point A, then BAT is equal to ;)
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According to the question, A circle with centre O, diameter AC and ∠ACB = 50° AT is a tangent to the circle at point A Since, angle in a semicircle is a right angle ∠CBA = 90°
By angle sum property of a triangle,
∠ACB + ∠CAB + ∠CBA = 180°
50° + ∠CAB + 90° = 180°
∠CAB = 40° … (1)
Since tangent to at any point on the circle is perpendicular to the radius through point of contact, We get, OA ⏊ AT ∠OAT = 90°
Answers & Comments
Explanation:
According to the question, A circle with centre O, diameter AC and ∠ACB = 50° AT is a tangent to the circle at point A Since, angle in a semicircle is a right angle ∠CBA = 90°
By angle sum property of a triangle,
∠ACB + ∠CAB + ∠CBA = 180°
50° + ∠CAB + 90° = 180°
∠CAB = 40° … (1)
Since tangent to at any point on the circle is perpendicular to the radius through point of contact, We get, OA ⏊ AT ∠OAT = 90°
∠OAT + ∠BAT = 90°
∠CAT + ∠BAT = 90°
40° + ∠BAT = 90°
BAT = 50⁰
Answer:
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Explanation:
BAT = 130°