Answer:
A = 151.4°, C = 90.6°, c = 30.8
Explanation:
Given a triangle with side lengths a = 14, b = 32, and angle B = 29°, we will use the Law of Cosines and the Law of Sines to solve for the remaining parts of the triangle.
1. Solve for side c using the Law of Cosines:
\[ c = \sqrt{a^2 + b^2 - 2ab\cos(B)} \]
2. Solve for angle A using the Law of Sines:
\[ \frac{\sin(A)}{a} = \frac{\sin(B)}{b} \]
\[ A = \arcsin\left(\frac{a\sin(B)}{b}\right) \]
3. Calculate angle C knowing that the sum of angles in a triangle is 180°:
\[ C = 180° - A - B \]
Let's compute the values.
Answer:
A = 19.0°, C = 132.0°, c = 20.9
Explanation:
After recalculating with the correct process, the angles and side of the triangle are as follows:
- Angle A is approximately 19.0 degrees.
- Angle C is approximately 132.0 degrees.
- Side c is approximately 20.9 units long.
These values are rounded to the nearest tenth as requested.
