Parts of a Triangle: Is it possible...?

To determine the angle $\angle B$ in triangle ABC with given $\angle A = 20^\circ$ and $\angle B > 90^\circ$, consider these facts:

The sum of all angles in any triangle is $180^\circ$.

With $\angle A = 20^\circ$, and knowing that $\angle B$ should be greater than $90^\circ$, mathematically, the sum of $\angle B$ and $\angle C$ should be $160^\circ$. However, without specific value for $\angle C$, multiple combinations of $\angle B$ and $\angle C$ that satisfy this condition exist.

To determine a unique value for $\angle B$, more information about $\angle C$ or any other angles or conditions is needed.

Thus, with the current information, it is not possible to calculate an exact measure for $\angle B$. Hence, the answer is No.

To solve this problem, we need to determine the measure of angle $∢C$ in the right triangle $\triangle ABC$ where angle $∢A = 20°$.

Since $\triangle ABC$ is a right triangle, we know that one angle, $∢B$, is $90°$. Hence, the other two angles, $∢A$ and $∢C$, must sum to $90°$ as well.

We are given that $∢A = 20°$. Therefore, we can set up the equation:$∢A + ∢C = 90°$

Substitute the given value of $∢A$ into the equation:
$20° + ∢C = 90°$

To solve for $∢C$, subtract $20°$ from both sides:
$∢C = 90° - 20°$

Thus, we find that:
$∢C = 70°$

Therefore, the size of angle $∢C$ is $\textbf{70°}$.

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 $\angle A = 40^\circ$.