Question : ABC is an isosceles triangle such that AB = AC, $\angle$ABC = 55°, and AD is the median to the base BC. Find the measure of $\angle$BAD.
Option 1: 50°
Option 2: 55°
Option 3: 35°
Option 4: 90°
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Correct Answer: 35°
Solution : Given, ABC is an isosceles triangle with AB = AC $\angle$ ABC = 55° AD is the median to the base BC which bisects $\angle$BAC By angle sum property in $\triangle$ABC, $\angle$BAC + $\angle$ABC + $\angle$ACB = 180° or, $\angle$BAC + 55° + 55° = 180° or, $\angle$ BAC = 70° So, $\angle$BAD = $\angle$CAD = $\frac{1}{2}\angle$BAC or, $\angle$BAD= 35° Hence, the correct answer is 35°.
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Question : In an isosceles $\triangle ABC$, $AB = AC$, $XY || BC$. If $\angle A=30°$, then $\angle BXY$?
Option 1: 75°
Option 2: 30°
Option 3: 150°
Option 4: 105°
Question : The vertical angle $\angle A$ of an isosceles $\triangle ABC$ is three times the angle B on it. The measure of the $\angle A$ is:
Option 1: 90°
Option 2: 108°
Option 3: 100°
Option 4: 36°
Question : In $\triangle$ABC, AB = $a - b$, AC = $\sqrt{a^2+b^2}$, and BC = $\sqrt{2ab}$, then find angle B.
Option 1: 60°
Option 3: 90°
Option 4: 45°
Question : In a triangle ABC, AB = 6$\sqrt{3}$ cm, AC = 12 cm and BC = 6 cm. Then the measure of $\angle B$ is equal to:
Option 2: 45°
Option 3: 70°
Option 4: 60°
Question : In $\triangle$ABC, $\angle$B = 35°, $\angle$C = 65° and the bisector of $\angle$BAC meets BC in D. Then $\angle$ADB is:
Option 1: 40°
Option 2: 75°
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