Question: Consider a triangle \( \triangle ABC \) like the one below. Suppose that \( b = 25 \), \( a =…

Consider a triangle \( \triangle ABC \) like the one below. Suppose that \( b = 25 \), \( a = 31 \), and \( B = 33^\circ \). (The figure is not drawn to scale.) Solve the triangle.

Carry your intermediate computations to at least four decimal places, and round your answers to the nearest tenth.

If no such triangle exists, enter “No solution.” If there is more than one solution, use the button labeled “or”.

[Diagram of triangle \( \triangle ABC \) with sides \( a \), \( b \), and \( c \) and angles \( A \), \( B \), and \( C \)]

\( A = \square^\circ \), \( C = \square^\circ \), \( c = \square \)

Solution

The problem involves solving a triangle \(ABC\) with the given values \(b = 25\), \(a = 31\), and \(B = 33^\circ\). We’ll use the law of sines and the law of cosines to find the missing angles and side. First, use the law of sines to find angle \(A\): \[ \frac{a}{\sin A} = \frac{b}{\sin B} \] \[ \frac{31}{\sin A} = \frac{25}{\sin 33^\circ} \] \[ \sin A = \frac{31 \cdot \sin 33^\circ}{25} \] Calculate \(\sin A\): \[ \sin A = \frac{31 \cdot 0.5446}{25} \] \[ \sin A = 0.6750 \] Find angle \(A\) using the inverse sine function: \[ A = \sin^{-1}(0.6750) \approx 42.7^\circ \] Now find angle \(C\) using the angle sum property of a triangle: \[ C = 180^\circ - A - B \] \[ C = 180^\circ - 42.7^\circ - 33^\circ \approx 104.3^\circ \] Now use the law of sines again to find side \(c\): \[ \frac{c}{\sin C} = \frac{a}{\sin A} \] \[ c = \frac{31 \cdot \sin 104.3^\circ}{\sin 42.7^\circ} \] Calculate \(c\): \[ c = \frac{31 \cdot 0.97236}{0.6781} \] \[ c \approx 44.5 \] The solutions for the triangle are: \[ \begin{cases} A \approx 42.7^\circ & \\ C \approx 104.3^\circ & \\ c \approx 44.5 & \end{cases} \]