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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Important Questions from Lines and Angles
If angles of a triangle are in the ration of 2 : 3 : 4, then the measure of the smallest angle is:
In the triangle, if AB = AC and ∠ABC = 72°, then ∠BAC is:
The angles of a triangle are (8x - 15)°,(6x - 11)° and ( 4x – 10)°. What is the value of x ?
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In ΔABC, D is a point on side BC such that ∠ADC = 2∠BAD. If ∠A = 80° and ∠C = 38°, then what is the measure of ∠ADB?
