Question

Consider a triangle ABC like the one below. Suppose that b=27, c=66, and B=130º. (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".

Answer 1

Answer:

The remaining dimensions of the triangle are $A \approx 31.7368^{\circ}$, $C \approx 18.2632^{\circ}$ and $a \approx 45.3201$.

Step-by-step explanation:

As angle B is an obtuse angle, Angle C can be obtained by means of the Law of Sine:

$\frac{b}{\sin B}=\frac{c}{\sin C}$

$\sin C = \frac{b}{c}\cdot \sin B$

$C = \sin^{-1}\left(\frac{b}{c}\cdot \sin B \right)$

Where:

$b$, $c$ - Measures of triangle sides, dimensionless.

$B$, $C$ - Measures of angles, measured in degrees.

If $b = 27$, $c = 66$ and $B =130^{\circ}$, then:

$C = \sin^{-1}\left(\frac{27}{66}\cdot \sin 130^{\circ} \right)$

$C \approx 18.2632^{\circ}$

Given that sum of internal angles in triangles equals to 180º, the angle A is now determined:

$A = 180^{\circ}-B-C$

$A = 180^{\circ}-130^{\circ}-18.2632^{\circ}$

$A \approx 31.7368^{\circ}$

Lastly, the length of the side $a$ is calculated by Law of Cosine:

$a = \sqrt{b^{2}+c^{2}-2\cdot b\cdot c\cdot \cos A}$

$a =\sqrt{27^{2}+66^{2}-2\cdot (27)\cdot (66)\cdot \cos 31.7368^{\circ}}$

$a \approx 45.3201$

The remaining dimensions of the triangle are $A \approx 31.7368^{\circ}$, $C \approx 18.2632^{\circ}$ and $a \approx 45.3201$.