Question : In a $\triangle \mathrm{XYZ}, \mathrm{XO}$ is the median and $\mathrm{XO}=\frac{1}{2} \mathrm{YZ}$. If $\angle \mathrm{YXO}=30^{\circ}$, then what is the value of $\angle \mathrm{XYZ}$?
Option 1: 15°
Option 2: 90°
Option 3: 30°
Option 4: 60°
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Correct Answer: 30°
Solution :
Since $XO$ is the median, $YO = OZ = \frac{YZ}{2}$ Given $XO = \frac{YZ}{2}$, ⇒ $YO = XO$ ⇒ $\triangle XYO$ is an isosceles triangle. Given, $\angle YXO = 30°$ Since angles opposite to equal sides of an isosceles triangle are equal, ⇒ $\angle XYO = 30°$ ⇒ $\angle XYZ = 30°$ Hence, the correct answer is 30°.
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Question : In $\triangle \mathrm{ABC}$, $\angle \mathrm{ABC} = 90^{\circ}$, $\mathrm{BP}$ is drawn perpendicular to $\mathrm{AC}$. If $\angle \mathrm{BAP} = 50^{\circ},$ what is the value of $\angle \mathrm{PBC}?$
Question : $\triangle \mathrm{XYZ} \sim \triangle \mathrm{GST}$ and $\mathrm{XY}: \mathrm{GS}=2: 3, \mathrm{XV}$ is the median to the side $\mathrm{YZ}$, and $\mathrm{GD}$ is the median to the side ST. The value of $\left(\frac{\mathrm{YV}}{\mathrm{SD}}\right)^2$ is:
Question : If $\triangle \mathrm{BPQ} \cong \triangle \mathrm{ASR}$ and $\angle \mathrm{A}=\frac{1}{3} \angle \mathrm{R}=\angle \mathrm{S}$, then find $\angle \mathrm{Q}$. (All angles are in degrees).
Question : In $\triangle$XYZ, P is the incentre of the $\triangle$XYZ. If $\angle$XYZ = 50°, then what is the value of $\angle$XPZ?
Question : In $\triangle$ABC, if the median AD = $\frac{1}{2}$BC, then $\angle$BAC is equal to:
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