Using the information given in the figure match the following :

y cm

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x^{\circ}

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90°- x°2

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\frac{y}{2}

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Solution

$∠ A O B = ∠ F O E = x^{\circ}$ ,,,,, (given)

We know that equal chords subtend equal angles at the centre.
Hence, FE = AB = y cm and $∠ C O D = ∠ A O B = x^{\circ}$

GF = $\frac{y}{2}$ cm .... (perpendicular from the centre bisects the chord)

$∠ O C D = ∠ O E G = 90^{\circ} - \frac{x^{\circ}}{2}$ .... (sum of angles of a triangle = 180 $^{\circ}$

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