Question:
If the angles of a triangle \(ABC\) are in the ratio \(1:2:3\), then the corresponding sides are in the ratio
If the angles of a triangle \(ABC\) are in the ratio \(1:2:3\), then the corresponding sides are in the ratio
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If the angles of a triangle are known, use the sine rule: sides are proportional to the sines of their opposite angles.
Updated On: Jun 15, 2026
- \(\sqrt3:2:1\)
- \(\sqrt3:1:2\)
- \(1:\sqrt3:2\)
- \(1:2:\sqrt3\)
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The Correct Option is C
Solution and Explanation
Step 1: Find the angles of the triangle.
Let the angles be
\[ x,\;2x,\;3x \]
Since the sum of angles of a triangle is \(180^\circ\),
\[ x+2x+3x=180^\circ \]
\[ 6x=180^\circ \]
\[ x=30^\circ \]
Therefore, the angles are
\[ 30^\circ,\;60^\circ,\;90^\circ \]
Step 2: Use the sine rule.
In a triangle, sides are proportional to the sines of opposite angles.
Thus,
\[ a:b:c = \sin30^\circ:\sin60^\circ:\sin90^\circ \]
Using standard values,
\[ \sin30^\circ=\frac12 \] \[ \sin60^\circ=\frac{\sqrt3}{2} \] \[ \sin90^\circ=1 \]
Hence,
\[ a:b:c = \frac12:\frac{\sqrt3}{2}:1 \]
Multiplying throughout by \(2\),
\[ a:b:c = 1:\sqrt3:2 \]
Step 3: Final conclusion.
Therefore, the corresponding sides are in the ratio
\[ \boxed{1:\sqrt3:2} \]
Let the angles be
\[ x,\;2x,\;3x \]
Since the sum of angles of a triangle is \(180^\circ\),
\[ x+2x+3x=180^\circ \]
\[ 6x=180^\circ \]
\[ x=30^\circ \]
Therefore, the angles are
\[ 30^\circ,\;60^\circ,\;90^\circ \]
Step 2: Use the sine rule.
In a triangle, sides are proportional to the sines of opposite angles.
Thus,
\[ a:b:c = \sin30^\circ:\sin60^\circ:\sin90^\circ \]
Using standard values,
\[ \sin30^\circ=\frac12 \] \[ \sin60^\circ=\frac{\sqrt3}{2} \] \[ \sin90^\circ=1 \]
Hence,
\[ a:b:c = \frac12:\frac{\sqrt3}{2}:1 \]
Multiplying throughout by \(2\),
\[ a:b:c = 1:\sqrt3:2 \]
Step 3: Final conclusion.
Therefore, the corresponding sides are in the ratio
\[ \boxed{1:\sqrt3:2} \]
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