In quadrilateral ABOC, the sum of all interior angles is 360^°. Since OBA + OCA = 90^° + 90^° = 180^°, the opposite angles BOC and BAC are supplementary, which yields BOC = 180^° - 118^° = 62^°.

In quadrilateral ABOC, the sum of all interior angles is 360^°. Since OBA + OCA = 90^° + 90^° = 180^°, the opposite angles BOC and BAC are supplementary, which yields BOC = 180^° - 118^° = 62^°.

The sum of angles in OBC is 180^°, which means BOC + 2 OCB = 180^°. Substituting BOC = 62^°, we find 2 OCB = 118^°, giving OCB = 59^°. Note that OCB is always simply half of BAC.

$AB$ and $AC$ are tangents from an external point $A$ to a circle with center $O$ . Tangents $AB$ and $AC$ touch the circle at $B$ and $C$ respectively such that $\angle BAC = 118^\circ$ . Find the measure of $\angle OCB$ . (a) $59^\circ$ (b) $62^\circ$ (c) $60^\circ$ (d) $54^\circ$ 28. एक समलम्ब चतुर्भुज $ABCD$ में, $DC \parallel AB$ , $AB = 16\text{ सेमी}$ और $DC = 11.2\text{ सेमी}$ है। इसके विकर्ण के मध्य बिन्दुओं को जोड़ने वाले रेखाखण्ड की लम्बाई (सेमी में) ज्ञात करें। In a trapezium $ABCD$ , $DC \parallel AB$ , $AB = 16\text{ cm}$ and $DC = 11.2\text{ cm}$ . Find the length (in $\text{cm}$ ) of the line segment joining the mid-points of its diagonal. (a) $2.4$ (b) $1.2$ (c) $1.8$ (d) $2.8$ 29. $O$ केन्द्र वाले तथा $10\text{ सेमी}$ त्रिज्या वाले एक वृत्त की एक $12\text{ सेमी}$ लम्बी जीवा $AB$ है। $A$ और $B$ पर खींची गई स्पर्शरेखाएँ $P$ पर प्रतिच्छेदित करती हैं। $OP$ की लम्बाई ज्ञात करें। A $12\text{ cm}$ long chord $AB$ of a circle with center $O$ and radius $10\text{ cm}$ . The tangents drawn at $A$ and $B$ intersect at $P$ . Find the length of $OP$ . (a) $12\text{ cm}$ (b) $10\text{ cm}$ (c) $2\sqrt{61}\text{ cm}$ (d) $12.5\text{ cm}$ 30. किसी वृत्त की दो बराबर समानान्तर जीवाओं के बीच की दूरी $10\text{ सेमी}$ है। यदि जीवाओं की लम्बाई $24\text{ सेमी}$ हो, तो त्रिज्या की लम्बाई ज्ञात करें। The distance between two equal parallel chords of a circle is $10\text{ cm}$ . If the length of the chords is $24\text{ cm}$ , then find the length of the radius. (a) $17\text{ cm}$ (b) $26\text{ cm}$ (c) $13\text{ cm}$ (d) $2\sqrt{61}\text{ cm}$ 31. $\triangle ABC$ में, माध्यिकाएँ $BE$ और $CF$ एक-दूसरे के लम्बवत हैं और बिन्दु $M$ पर प्रतिच्छेदित करती हैं। यदि $BE = 9\text{ सेमी}$ और $CF = 13\text{ सेमी}$ है, तो $\triangle ABC$ का क्षेत्रफल ज्ञात करें।

Answer: (a) $59^\circ$

Step-by-step solution

Step 1: Use tangent-radius perpendicularity

By the tangent-radius theorem, the radius through the point of contact is perpendicular to the tangent. Therefore, $OB \perp AB$ and $OC \perp AC$ , giving $\angle OBA = 90^\circ$ and $\angle OCA = 90^\circ$ .

Step 2: Find angle BOC at the center

In quadrilateral $ABOC$ , the sum of all interior angles is $360^\circ$ . Since $\angle OBA + \angle OCA = 90^\circ + 90^\circ = 180^\circ$ , the opposite angles $\angle BOC$ and $\angle BAC$ are supplementary, which yields $\angle BOC = 180^\circ - 118^\circ = 62^\circ$ .

Step 3: Apply isosceles triangle properties in triangle OBC

Since segments $OB$ and $OC$ are both radii of the circle, triangle $OBC$ is an isosceles triangle with base $BC$ . Thus, the base angles $\angle OBC$ and $\angle OCB$ are equal in measure.

Step 4: Calculate the measure of angle OCB

The sum of angles in $\triangle OBC$ is $180^\circ$ , which means $\angle BOC + 2\angle OCB = 180^\circ$ . Substituting $\angle BOC = 62^\circ$ , we find $2\angle OCB = 118^\circ$ , giving $\angle OCB = 59^\circ$ . Note that $\angle OCB$ is always simply half of $\angle BAC$ .

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