The sum of interior angles in ABC is 180^°. Subtracting BAC = 70^° leaves 110^° for the two equal base angles, so each base angle ABC = (110^°)/(2) = 55^°.

The side $BC$ of $\triangle ABC$ is produced to $D$ . The bisectors of $\angle ABC$ and $\angle ACD$ meet at point $E$ . If $AB = AC$ and $\angle BEC = 35^\circ$ , then find the measure of $\angle ABC$ . (SSC CHSL 2020) (a) $75^\circ$ (b) $35^\circ$ (c) $45^\circ$ (d) $55^\circ$ 26. $\triangle ABC$ की भुजा $AB$ और $AC$ , बिन्दु $D$ और $E$ तक बढ़ाई जाती हैं। इस प्रकार बने बाह्य कोणों के समद्विभाजक एक-दूसरे को बिन्दु $I$ पर प्रतिच्छेदित करते हैं। यदि $\angle ACB = 66^\circ$ है और $\angle ABC = 44^\circ$ है, तो $\angle BIC$ की माप (डिग्री में) ज्ञात करें। The sides $AB$ and $AC$ of $\triangle ABC$ are produced to the points $D$ and $E$ . The bisectors of the exterior angles so formed intersect each other at point $I$ . If $\angle ACB = 66^\circ$ and $\angle ABC = 44^\circ$ , find the measure (in degrees) of $\angle BIC$ . (SSC CHSL 2020) (a) 52 (b) 48 (c) 50 (d) 55 27. $AB$ और $AC$ , बाह्य बिन्दु $A$ से केन्द्र $O$ वाले वृत्त की स्पर्श रेखाएँ हैं। स्पर्श रेखाएँ $AB$ और $AC$ क्रमशः बिन्दु $B$ और $C$ पर वृत्त को इस प्रकार स्पर्श करती हैं कि $\angle BAC = 118^\circ$ है। $\angle OCB$ की माप ज्ञात करें। $AB$ and $AC$ are tangents from an external point $A$ to a circle with center $O$ . Tangents $AB$ and $AC$ touch the circle at points $B$ and $C$ respectively such that $\angle BAC = 118^\circ$ . Find the measure of $\angle OCB$ .

Answer: 55°

Step-by-step solution

Step 1: Relate $\angle BEC$ to $\angle BAC$

In any $\triangle ABC$ , the interior bisector of $\angle B$ and the exterior bisector of $\angle C$ meet at the excenter $E$ , which satisfies the geometric property $\angle BEC = \frac{1}{2}\angle BAC$ . Since we are given that $\angle BEC = 35^\circ$ , multiplying by 2 gives $\angle BAC = 70^\circ$ .

Step 2: Use the Isosceles Triangle Property

We are given that $AB = AC$ , which makes $\triangle ABC$ an isosceles triangle with vertex $A$ . Angles opposite to equal sides are equal, so $\angle ABC = \angle ACB$ .

Step 3: Calculate the measure of $\angle ABC$

The sum of interior angles in $\triangle ABC$ is $180^\circ$ . Subtracting $\angle BAC = 70^\circ$ leaves $110^\circ$ for the two equal base angles, so each base angle $\angle ABC = \frac{110^\circ}{2} = 55^\circ$ .

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Step-by-step solution

Step 1: Relate ∠BEC\angle BEC∠BEC to ∠BAC\angle BAC∠BAC

Step 2: Use the Isosceles Triangle Property

Step 3: Calculate the measure of ∠ABC\angle ABC∠ABC

Step 1: Relate $\angle BEC$ to $\angle BAC$

Step 3: Calculate the measure of $\angle ABC$