Ab is a Chord of a Circle with Centre O , Aoc is a Diameter and at is the Tangent at a as Shown in Fig . 10.70. Prove that ∠ Bat = ∠ Acb. - Mathematics

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Short Note

AB is a chord of a circle with centre O , AOC  is a diameter and AT is the tangent at A as shown in Fig . 10.70. Prove that \[\angle\]BAT = \[\angle\] ACB

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Solution

In the given figure, 
AC is the diameter.
So, 

\[\angle CBA = 90^o\](Angle formed by the diameter on the circle is 90º)
AT is the tangent at point A. 
Thus, 
\[\angle CAT = 90^o\]
In ∆ABC,
\[\angle BCA + \angle BAC + \angle CBA = 180^o \left( \text{Angle sum property} \right)\]
\[ \Rightarrow \angle BCA + 90^o + \angle CAT - \angle BAT = 180^o \]
\[ \Rightarrow \angle BCA + 90^o + 90 - \angle BAT = 180^o \]
\[ \Rightarrow \angle BCA = \angle BAT\]

Hence Proved

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Chapter 8: Circles - Exercise 8.2 [Page 39]

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RD Sharma Class 10 Maths
Chapter 8 Circles
Exercise 8.2 | Q 41 | Page 39

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