Let ABC be a triangle and let D be the mid-po | vedclass

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(A) Let $\angle CAD = 	heta_2$ and $\angle BAD = 	heta_1$. We are given $\cot 	heta_2 : \cot 	heta_1 = 2 : 1$,so $\cot 	heta_2 = 2 \cot 	heta_1$

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$90$

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(A) Let $\angle CAD = 	heta_2$ and $\angle BAD = 	heta_1$. We are given $\cot 	heta_2 : \cot 	heta_1 = 2 : 1$,so $\cot 	heta_2 = 2 \cot 	heta_1$.Using the cotangent rule in $	riangle ADC$ and $	riangle ADB$ with $AD$ as a common side and $BD = DC = a/2$:$\cot 	heta_2 = \frac{AD^2 + b^2 - (a/2)^2}{2 \cdot AD \cdot (a/2) \cdot \sin 	heta_2} \cdot \sin 	heta_2 = \frac{AD^2 + b^2 - a^2/4}{2 \cdot 	ext{Area}(	riangle ADC)}$Since $	ext{Area}(	riangle ADC) = 	ext{Area}(	riangle ADB)$,the condition $\cot 	heta_2 = 2 \cot 	heta_1$ implies:$AD^2 + b^2 - a^2/4 = 2(AD^2 + c^2 - a^2/4)$$AD^2 = b^2 - 2c^2 + a^2/4$Using Apollonius theorem,$AD^2 = \frac{2b^2 + 2c^2 - a^2}{4}$.Equating the two expressions for $AD^2$:$\frac{2b^2 + 2c^2 - a^2}{4} = b^2 - 2c^2 + \frac{a^2}{4}$$2b^2 + 2c^2 - a^2 = 4b^2 - 8c^2 + a^2$$10c^2 = 2b^2 + 2a^2 \Rightarrow b^2 + a^2 = 5c^2$.In $	riangle BGA$,by the Law of Cosines:$\cos(\angle BGA) = \frac{AG^2 + BG^2 - AB^2}{2 \cdot AG \cdot BG}$.Using $AG = \frac{2}{3}m_a$ and $BG = \frac{2}{3}m_b$,where $m_a^2 = \frac{2b^2+2c^2-a^2}{4}$ and $m_b^2 = \frac{2a^2+2c^2-b^2}{4}$,and substituting $a^2+b^2=5c^2$,we find $\cos(\angle BGA) = 0$,so $\angle BGA = 90^{\circ}$.

Let $ABC$ be a triangle and let $D$ be the mid-point of $BC$. Suppose $\cot (\angle CAD) : \cot (\angle BAD) = 2 : 1$. If $G$ is the centroid of $\triangle ABC$,then the measure of $\angle BGA$ is (in $^{\circ}$)

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