In a triangle ABC, if 2 ∠A = 3 ∠B = 6 ∠C, then what is ∠A + ∠C equal to?
120°
Let's solve this geometry problem about a triangle ABC where we are given a specific relationship between its angles. We are told that 2 times angle A is equal to 3 times angle B, which is also equal to 6 times angle C. Our goal is to find the sum of angle A and angle C.
Understanding the Given Relationship
The problem provides the relationship: \(2 \angle A = 3 \angle B = 6 \angle C\).
This means all three expressions are equal to some common value. Let's call this common value \(k\).
\(2 \angle A = k\)
\(3 \angle B = k\)
\(6 \angle C = k\)
Expressing Angles in Terms of a Common Variable
From the relationships above, we can express each angle in terms of \(k\):
- \(\angle A = \frac{k}{2}\)
- \(\angle B = \frac{k}{3}\)
- \(\angle C = \frac{k}{6}\)
Using the Angle Sum Property of a Triangle
We know that the sum of the interior angles in any triangle is always 180 degrees. For triangle ABC, this means:
\(\angle A + \angle B + \angle C = 180^\circ\)
Solving for the Common Variable \(k\)
Now, substitute the expressions for \(\angle A\), \(\angle B\), and \(\angle C\) in terms of \(k\) into the angle sum equation:
\(\frac{k}{2} + \frac{k}{3} + \frac{k}{6} = 180^\circ\)
To solve for \(k\), find a common denominator for the fractions, which is 6.
\(\frac{3k}{6} + \frac{2k}{6} + \frac{k}{6} = 180^\circ\)
Combine the terms on the left side:
\(\frac{3k + 2k + k}{6} = 180^\circ\)
\(\frac{6k}{6} = 180^\circ\)
\(k = 180^\circ\)
Calculating the Angles
Now that we have the value of \(k\), we can find the measure of each angle:
- \(\angle A = \frac{k}{2} = \frac{180^\circ}{2} = 90^\circ\)
- \(\angle B = \frac{k}{3} = \frac{180^\circ}{3} = 60^\circ\)
- \(\angle C = \frac{k}{6} = \frac{180^\circ}{6} = 30^\circ\)
Verifying the Angles
Let's check if these angles satisfy the given relationship and the angle sum property:
- Check given relationship:
- \(2 \angle A = 2 \times 90^\circ = 180^\circ\)
- \(3 \angle B = 3 \times 60^\circ = 180^\circ\)
- \(6 \angle C = 6 \times 30^\circ = 180^\circ\)
- Check angle sum property:
- \(\angle A + \angle B + \angle C = 90^\circ + 60^\circ + 30^\circ = 180^\circ\)
Finding ∠A + ∠C
The question asks for the value of \(\angle A + \angle C\).
\(\angle A + \angle C = 90^\circ + 30^\circ = 120^\circ\)
Thus, the sum of angle A and angle C is \(120^\circ\).
Revision Table: Triangle Angles
| Concept | Description |
|---|---|
| Angle Sum Property | Sum of interior angles in a triangle is \(180^\circ\). |
| Given Relation | \(2 \angle A = 3 \angle B = 6 \angle C\) |
| Calculated Angles | \(\angle A = 90^\circ\), \(\angle B = 60^\circ\), \(\angle C = 30^\circ\) |
| Required Value | \(\angle A + \angle C = 120^\circ\) |
Additional Information: Types of Triangles
Based on their angles, triangles can be classified:
- Acute Triangle: All three angles are less than \(90^\circ\).
- Right Triangle: One angle is exactly \(90^\circ\).
- Obtuse Triangle: One angle is greater than \(90^\circ\).
In this problem, we found \(\angle A = 90^\circ\), \(\angle B = 60^\circ\), and \(\angle C = 30^\circ\). Since one angle is \(90^\circ\), the triangle ABC is a right triangle.
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