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Standard IX

Mathematics

Chord

O is the cent...

Question

O is the centre of the circle. If $∠$ BOA = 90° and $∠$ COA = 110°, find $∠$ BAC.

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Solution

. Given : A circle with centre O and $∠$ BOA = 90°, $∠$ AOC = 110°.

To find : $∠$ BAC
Proof : $∠$ AOB + $∠$ AOC + $∠$ BOC = 360°
$\Rightarrow$ 90° + 110° + $∠$ BOC = 360°
$\Rightarrow$ $∠$ BOC = 360° – 90° – 110°
$\Rightarrow$ $∠$ BOC = 160°

But, arc $ˆ B C$ subtends $∠$ BOC at the centre and $∠$ BAC at the remaining part of the circle.

$\Rightarrow ∠ B A C = \frac{1}{2} ∠ B O C ∴ ∠ B A C = \frac{1}{2} (160 °) = 80 °$

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O is the centre of the circle. If ∠BOA = 90° and ∠COA = 110°, find ∠BAC.

O is the centre of the circle. If $∠$ BOA = 90° and $∠$ COA = 110°, find $∠$ BAC.