Question : In $\triangle \mathrm{STU}, \mathrm{SX}$ is the median on $\mathrm{TU}$. If $\mathrm{SX}=\mathrm{TX}$, then what is the value of $\angle \mathrm{TSU}$?
Option 1: 75°
Option 2: 45°
Option 3: 60°
Option 4: 90°
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Correct Answer: 90°
Solution : In $\triangle$STU, SX is the median. So, TX = XU (median property) SX = TX (given) So, $\triangle$STX is an isosceles triangle. Let $\angle$STX = $\angle$TSX = $\theta$ ⇒ $\angle$SXU = $2\theta$ (exterior angle) Similarly in $\triangle$SXU, SX = XU $\triangle$SXU is an isosceles triangle. Let $\angle$SUX = $\angle$USX = $\alpha$ ⇒ $\angle$SXT = $2\alpha$ (exterior angle) Now, $2\theta + 2\alpha = 180° $ (Linear pair) $⇒\theta + \alpha = 90°$ So, $\angle$TSU $= \theta +\alpha = 90°$ Hence, the correct answer is 90°.
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Question : In $\triangle \mathrm{PQR}, \angle \mathrm{P}=46^{\circ}$ and $\angle \mathrm{R}=64^{\circ}$.If $\triangle \mathrm{PQR}$ is similar to $\triangle \mathrm{ABC}$ and in correspondence, then what is the value of $\angle \mathrm{B}$?
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