Let's call the angles at vertices $A$, $B$, and $C$ as $\angle A$, $\angle B$, and $\angle C$, respectively.

Step 1: Express relations between the angles.

From the problem:

• $\angle A = 2 \times \angle B$ (Angle at $A$ is twice that of $B$).
• $\angle B = \angle C + 16^\circ$ (Angle at $B$ is 16° more than that at $C$).

Step 2: Set up the equation for the sum of the angles.

In any triangle, the sum of the interior angles is always $180^\circ$: $\angle A + \angle B + \angle C = 180^\circ$

Step 3: Substitute the relations into the sum equation.

We already have $\angle A = 2 \times \angle B$ and $\angle B = \angle C + 16^\circ$. We can substitute $\angle C$ as $\angle B - 16^\circ$ into the equation.

Now, substituting into the sum of angles: $2 \times \angle B + \angle B + (\angle B - 16^\circ) = 180^\circ$

Step 4: Simplify the equation.

$2\angle B + \angle B + \angle B - 16^\circ = 180^\circ$ $4\angle B - 16^\circ = 180^\circ$ $4\angle B = 196^\circ$ $\angle B = \frac{196^\circ}{4} = 49^\circ$

Step 5: Calculate $\angle A$.

Now that we know $\angle B = 49^\circ$, we can calculate $\angle A$: $\angle A = 2 \times \angle B = 2 \times 49^\circ = 98^\circ$

Step 6: Calculate $\angle C$.

Using $\angle B = \angle C + 16^\circ$, we find: $\angle C = \angle B - 16^\circ = 49^\circ - 16^\circ = 33^\circ$

Thus, the angles in the triangle are:

• $\angle A = 98^\circ$
• $\angle B = 49^\circ$
• $\angle C = 33^\circ$

The value of $\angle A$ is 98°.

Would you like further details or have any questions?

Here are five related questions:

1. How do you solve for the angles of an isosceles triangle?
2. What is the sum of the interior angles in a quadrilateral?
3. Can you prove why the sum of angles in any triangle is 180°?
4. How does the exterior angle theorem apply in triangles?
5. What is the difference between an acute and obtuse triangle?

Tip: Remember, the sum of the interior angles of any polygon is $(n - 2) \times 180^\circ$, where $n$ is the number of sides.