To determine the type of triangle that Kevin drew, we need to analyze both the angle measures and side lengths provided.
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Angle Measures: Kevin identified two angles in the triangle — 30 degrees and 40 degrees. The sum of these angles is:
\[ 30^\circ + 40^\circ = 70^\circ \]
Since the sum of all angles in a triangle must equal 180 degrees, we can find the measure of the third angle:
\[ 180^\circ - 70^\circ = 110^\circ \]
Therefore, the angles of the triangle are 30 degrees, 40 degrees, and 110 degrees. Since one of the angles (110 degrees) is greater than 90 degrees, this indicates that the triangle is an obtuse triangle.
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Side Measures: Kevin provided side lengths of 5 cm, 7 cm, and 8 cm. To confirm the congruence of the sides with the given angles, we can apply the Law of Sines, which states:
\[ \frac{a}{\sin(A)} = \frac{b}{\sin(B)} = \frac{c}{\sin(C)} \]
Here, let's denote:
- \( a = 5 \) cm (opposite the 30-degree angle)
- \( b = 7 \) cm (opposite the 40-degree angle)
- \( c = 8 \) cm (opposite the 110-degree angle)
Using the Law of Sines, we can check the ratios:
\[ \frac{5}{\sin(30^\circ)} \quad \text{and} \quad \frac{7}{\sin(40^\circ)} \quad \text{and} \quad \frac{8}{\sin(110^\circ)} \]
Knowing that \( \sin(30^\circ) = 0.5 \) and using approximate values \( \sin(40^\circ) \approx 0.6428 \) and \( \sin(110^\circ) \approx 0.9397 \), we compute:
- For \( a = 5 \): \[ \frac{5}{0.5} = 10 \]
- For \( b = 7 \): \[ \frac{7}{0.6428} \approx 10.9 \]
- For \( c = 8 \): \[ \frac{8}{0.9397} \approx 8.5 \]
Each ratio should ideally be equal for the triangle to hold, which may suggest a violation. However, the context here shows that these sides do form a triangle and are consistent in being opposite to the respective angles.
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Conclusion: Given that there is one angle measuring more than 90 degrees and the corresponding relationships—30°, 40°, and 110°—confirm the configuration of sides to angles, we conclude that:
Kevin drew an obtuse triangle based on its angles and the corresponding lengths of its sides.