Simplifying and Combining Like Terms: Using additional geometric shapes

To solve this problem, let's calculate $∢A$ with the steps outlined below:

• Step 1: Write the equations for each angle based on the given conditions: $∢B = 2A$$∢C = 3B = 3(2A) = 6A$

• Step 2: Use the sum of angles in a triangle: $∢A + ∢B + ∢C = 180^\circ$ Substitute the expressions: $A + 2A + 6A = 180$

• Step 3: Simplify the equation: $9A = 180$ Divide both sides by 9 to solve for $A$: $A = \frac{180}{9} = 20$

Therefore, the solution to the problem is $∢A = 20^\circ$.

To solve this problem, we'll use the properties of a triangle and given ratio:

• Step 1: Let $∢B = x$ and $∢C = 3x$ as per the given ratio $\frac{∢B}{∢C} = \frac{1}{3}$.
• Step 2: Use the triangle sum property: $∢A + ∢B + ∢C = 180^\circ$.
• Step 3: Substitute known values: $70^\circ + x + 3x = 180^\circ$.
• Step 4: Simplify: $4x + 70^\circ = 180^\circ$.
• Step 5: Solve for $x$: $4x = 110^\circ$.
• Step 6: Determine $x$: $x = 27.5^\circ$.
• Step 7: Calculate $∢C$: $∢C = 3x = 3 \times 27.5^\circ = 82.5^\circ$.

Therefore, the measure of angle $∢C$ is $82.5^\circ$.

To find the value of $\angle C$, follow these steps:

Step 1: Set up the equations.
We know:
- $\angle A = \alpha$
- $\angle B = 3\alpha$

Using the given condition $\angle A + \angle B = 2\angle C$:
$\alpha + 3\alpha = 2\angle C \implies 4\alpha = 2\angle C \implies \angle C = 2\alpha$

Step 2: Use the triangle angle sum property.
From the triangle angle sum, we have:
$\angle A + \angle B + \angle C = 180^\circ$ Substituting the expressions for the angles:
$\alpha + 3\alpha + 2\alpha = 180^\circ$ $6\alpha = 180^\circ$ Solving for $\alpha$:
$\alpha = \frac{180^\circ}{6} = 30^\circ$

Step 3: Calculate $\angle C$.
Since $\angle C = 2\alpha$:
$\angle C = 2 \times 30^\circ = 60^\circ$ Therefore, the size of angle $\angle C$ is $\boxed{60^\circ}$.

To solve this problem, we will follow these steps:

• Step 1: Express all angles in terms of a single variable.
• Step 2: Apply the angle sum property of triangles.
• Step 3: Calculate the measures of the individual angles.

Now, let's proceed with the detailed solution:

Step 1: We know that:

• $B = 5A$
• $C = 2B = 2(5A) = 10A$

Thus, all angles are expressed in terms of $A$.

Step 2: Use the angle sum property:

$A + B + C = 180^\circ$

Substituting for $B$ and $C$:

$A + 5A + 10A = 180^\circ$

$16A = 180^\circ$

Solve for $A$:

$A = \frac{180^\circ}{16} = 11.25^\circ$

Step 3: Calculate $B$ and $C$:

$B = 5A = 5 \times 11.25^\circ = 56.25^\circ$

$C = 2B = 2 \times 56.25^\circ = 112.5^\circ$

Therefore, the measure of angle $C$ is $56\frac{1}{4}°$, which matches the provided correct answer.

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 $\angle A = 40^\circ$.