In ∆ABC, 2∠A = 3∠B = 6∠C. What is the value of the largest angle among these three angles?
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The problem provides a relationship between the three angles of a triangle ▵ABC. The relationship is given as $2\angle A = 3\angle B = 6\angle C$. We need to find the value of the largest angle among these three angles.
In any triangle, the sum of the interior angles is always $180^\circ$. This is known as the Angle Sum Property of a triangle. So, we have:
We are given the relationship: $2\angle A = 3\angle B = 6\angle C$. Let's assume that this common value is equal to some constant, say $k$.
So, we have:
Now, we can substitute these expressions for $\angle A$, $\angle B$, and $\angle C$ into the angle sum property equation:
$\frac{k}{2} + \frac{k}{3} + \frac{k}{6} = 180^\circ$
To solve for $k$, we 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:
$\frac{3k + 2k + k}{6} = 180^\circ$
$\frac{6k}{6} = 180^\circ$
$k = 180^\circ$
Now that we have the value of $k$, we can find the measure of each angle:
Let's check if the sum of these angles is $180^\circ$: $90^\circ + 60^\circ + 30^\circ = 180^\circ$. This confirms our calculations are correct.
The three angles of the triangle are $90^\circ$, $60^\circ$, and $30^\circ$. We need to find the largest angle among these three angles.
Comparing the values: $90^\circ > 60^\circ > 30^\circ$.
The largest angle is $90^\circ$.
Let's summarize the angles in a table:
| Angle | Value |
|---|---|
| $\angle A$ | $90^\circ$ |
| $\angle B$ | $60^\circ$ |
| $\angle C$ | $30^\circ$ |
The largest angle is indeed $90^\circ$.
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