Sum and Difference of Angles: Is it possible...?

To solve for $\angle A$ in triangle $\triangle ABC$, we proceed as follows:

• First, note that the sum of angles in any triangle is $180^\circ$. Therefore, $\angle A + \angle B + \angle C = 180^\circ$.
• We know that $\angle B = 3 \angle A$ and $\angle C = \frac{1}{2} \angle A$.
• Substitute these expressions into the triangle sum equation: $\angle A + 3\angle A + \frac{1}{2}\angle A = 180^\circ$.
• Combine like terms: $\angle A + 3\angle A + \frac{1}{2}\angle A = 4\angle A + \frac{1}{2}\angle A = \frac{9}{2}\angle A$.
• The equation becomes $\frac{9}{2} \angle A = 180^\circ$.
• To solve for $\angle A$, multiply both sides by $\frac{2}{9}$:
$\angle A = \frac{2}{9} \times 180^\circ = 40^\circ$.
• Check consistency: $\angle A = 40^\circ$ leads to $\angle B = 120^\circ$ and $\angle C = 20^\circ$.
• Verify that $\triangle ABC$ is consistent with being obtuse: Indeed, the triangle has $\angle B = 120^\circ$ which is greater than $90^\circ$, confirming the triangle is obtuse.

Therefore, it is possible to calculate $\angle A$, and the solution is $\angle A = 40^\circ$.