Problem

Determine the remaining sides and angles of the triangle $A B C$. \[ A=28.79^{\circ} \quad B=19.07^{\circ} \quad c=11.49 m \] \[ \mathrm{C}= \] \[ a \approx \square m \] m (Round to the nearest hundredth as needed.) (Round to the nearest hundredth as needed.)

Solution

Step 1 :Given that in a triangle, the sum of angles is 180 degrees, we can calculate the third angle C as \(180 - A - B = 180 - 28.79 - 19.07 = 132.14^{\circ}\).

Step 2 :We can then use the Law of Sines to find the remaining sides. The Law of Sines states that the ratio of the length of a side of a triangle to the sine of its opposite angle is the same for all three sides of the triangle. Therefore, we can write the following equations: \(\frac{a}{\sin A} = \frac{c}{\sin C}\) and \(\frac{b}{\sin B} = \frac{c}{\sin C}\).

Step 3 :Solving these equations for a and b, we get \(a = c \cdot \frac{\sin A}{\sin C} \approx 7.46 m\) and \(b = c \cdot \frac{\sin B}{\sin C} \approx 5.06 m\).

Step 4 :\(\boxed{C = 132.14^{\circ}, a \approx 7.46 m, b \approx 5.06 m}\)

From Solvely APP
Source: https://solvelyapp.com/problems/19592/

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