Question

Two chords AB and AC of a circle subtend angles equal to 90° and 150°,respectively at the centre. Find ∠BAC, if AB and AC lie on the opposite sides of the centre.

Answer 1

Answer:the question is wrong as two chords can't have similar name( A)

Step-by-step explanation:

Answer 2

$\large\underline{Diagram:-}$

$\large\underline{Questions:-}$

Two chords AB and AC of a circle subtend angles equal to 90° and 150°,respectively at the centre. Find ∠BAC, if AB and AC lie on the opposite sides of the centre..

$\large\underline{Given:-}$

• Two chords AB and AC of a circle subtend angles equal to 90° and 150°.

$\large\underline{To find:-}$

• Find ∠BAC, if AB and AC lie on the opposite sides of the centre ...?

$\large\underline{Solutions:-}$

In ∆BOA,

$\: \: \: \: \: OB \: \: = \: \: OA \: \: \: \: \: \: \: {(both \: \: are \: \: the \: \: radius \: \: of \: \: circle)}$

$\: \: \: \: \: \therefore \: \: \angle OAB \: \: = \: \: \angle OBA \: \: \: \: \: \: \: {(angle \: \: opposite \: \: to \: \: equal \: \: side \: \: are \: \: equal)}$

In ∆OAB,

$\: \: \: \: \: \angle OBA \: + \: \angle OAB \: + \: \angle AOB \: \: = \: \: {180} \degree \: \: \: \: \: \: {(by \: \: angle \: \: sum \: \: property \: \: of \: \: a \: \: triangle)}$

$\: \: \: \: \: \angle OBA \: + \: \angle OAB \: + \: {90} \degree \: \: = \: \: {180} \degree$

$\: \: \: \: \: {2} \: \angle OAB \: \: = \: \: {180} \degree \: - \: {90} \degree$

$\: \: \: \: \: {2} \: \angle OAB \: \: = \: \: {90} \degree$

$\: \: \: \: \: \angle OAB \: \: = \: \: \frac{{90} \degree}{2}$

$\: \: \: \: \: \angle OAB \: \: = \: \: {45} \degree$

Now, In ∆AOC,

$\: \: \: \: \: AO \: \: = \: \: OC \: \: \: \: \: \: \: {(both \: \: are \: \: the \: \: radius \: \: of \: \: circle)}$

$\: \: \: \: \: \therefore \: \: \angle OCA \: \: = \: \: \angle OAC \: \: \: \: \: \: \: {(angle \: \: opposite \: \: to \: \: equal \: \: side \: \: are \: \: equal)}$

Also,

$\: \: \: \: \: \angle AOC \: + \: \angle OAC \: + \: \angle OCA \: \: = \: \: {180} \degree \: \: \: \: \: \: {(by \: \: angle \: \: sum \: \: property \: \: of \: \: a \: \: triangle)}$

$\: \: \: \: \: {150} \degree \: + \: \angle OAC \: + \: \angle OCA \: \: = \: \: {180} \degree$

$\: \: \: \: \: {2} \: \angle OAC \: \: = \: \: {180} \degree \: - \: {150} \degree$

$\: \: \: \: \: {2} \: \angle OAC \: \: = \: \: {30} \degree$

$\: \: \: \: \: \angle OAC \: \: = \: \: \frac{{30} \degree}{2}$

$\: \: \: \: \: \angle OAB \: \: = \: \: {15} \degree$

$\: \: \: \: \: \therefore \: \: \angle BAC \: \: = \: \angle OAB \: + \: \angle OAC$

$\: \: \: \: \: \angle BAC \: \: = \: {45} \degree \: + \: {15} \degree$

$\: \: \: \: \: \angle BAC \: \: = \: {60} \degree$

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