In the Given Figure, O is the Centre of the Circle and Bcd is Tangent to It at C. Prove that ∠Bac + ∠Acd = 90°. - Mathematics

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Short Note

In the given figure, O is the centre of the circle and BCD is tangent to it at C. Prove that ∠BAC + ∠ACD = 90°.

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Solution

In the given figure, let us join D an A.

Consider `Δ OCA`. We have,

OC = OA (Radii of the same circle)

We know that angles opposite to equal sides of a triangle will be equal. Therefore,

`∠ OCA =∠ OAC ` …… (1)

It is clear from the figure that

`∠ DCA +∠ OCA=∠ OCD`

Now from (1)

`∠ DCA +∠OAC =∠ OCD`

Now as BD is tangent therefore `∠ OCD=90^o`

Therefore `∠ DCA +∠ OAC = 90^o`

From the figure we can see that `∠ OAC =∠ BAC`

`∠ DAC + ∠ BAC = 90^o`

Thus we have proved.

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Chapter 8: Circles - Exercise 8.2 [Page 41]

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RD Sharma Class 10 Maths
Chapter 8 Circles
Exercise 8.2 | Q 48 | Page 41

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