Solving Triangle Angles A B C
The problem requires finding the measures of the three angles in a triangle, $\angle A$, $\angle B$, and $\angle C$, given specific relationships between them. We know that the sum of the interior angles in any triangle is always $180^\circ$.
Given relationships:- $\angle A = 3\angle B$
- $\angle C = 2\angle B$
The sum of the angles in $\Delta ABC$ is $180^\circ$:
$ \angle A + \angle B + \angle C = 180^\circ $Substitute the given relationships into the equation to express everything in terms of $\angle B$:
$ (3\angle B) + \angle B + (2\angle B) = 180^\circ $Combine the terms involving $\angle B$:
$ 6\angle B = 180^\circ $Solve for $\angle B$:
$ \angle B = \frac{180^\circ}{6} $ $ \angle B = 30^\circ $Now, calculate $\angle A$ and $\angle C$ using the value of $\angle B$:
- $\angle A = 3\angle B = 3 \times 30^\circ = 90^\circ$
- $\angle C = 2\angle B = 2 \times 30^\circ = 60^\circ$
Therefore, the values of the angles are $\angle A = 90^\circ$, $\angle B = 30^\circ$, and $\angle C = 60^\circ$. This matches Option B.
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