Find the area of the triangle:

1. \( C = 110^\circ \), \( a = 6 \), \( b = 10 \)

2. \( B = 130^\circ \), \( a = 92 \), \( c = 30 \)

The area of the triangle with angles C = 110 degrees, side lengths a = 6, and b = 10 is approximately 24 square units.

The area of the triangle with angle B = 130 degrees, side lengths a = 92, and c = 30 cannot be determined as the given side lengths and angle are not compatible for a triangle.

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Answered by muhammadqadeer4187 | 2024-06-18

To find the area of a triangle with sides a and b and an included angle C, the formula based on the Law of Cosines is used: ½ × a × b × sin(C). For the first triangle with a=6, b=10, and C=110 degrees, the area can be calculated:
Area = ½ × 6 × 10 × sin(110°).
For the second triangle with a=92, c=30, and B=130 degrees, we first need to find the third side b using the Law of Cosines and then calculate the area. Given that c² = a² + b² - 2 a b*cos(B), we can find b and then use the previous area formula.
Keep in mind, to give the final answer to the proper number of significant figures, depending on how the calculation is performed.
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Answered by CharlesBronson | 2024-06-19

The area of the triangle with angles C = 110°, side lengths a = 6, and b = 10 is approximately 28.19 square units. For the triangle with angle B = 130°, side lengths a = 92, and c = 30, the area is approximately 1497.83 square units after applying Heron's formula. Both areas were calculated using trigonometric and geometric principles.
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Answered by muhammadqadeer4187 | 2024-12-29