Use the Law of Sines to solve for all possible triangles that satisfy the given conditions. Below, enter your answers so that ∠A1 is smaller than ∠A2.)

a = 71, b = 104, ∠A = 21°
∠B1 = ∠B2 =
∠C1 = ∠C2 =
c1 = c2 =

Respuesta :

Answer:

[tex]\angle B_{1} \approx 31.668^{\circ}[/tex], [tex]\angle B_{2} \approx 148.332^{\circ}[/tex]

[tex]\angle C_{1} \approx 127.332^{\circ}[/tex], [tex]\angle C_{2} \approx 10.668^{\circ}[/tex]

[tex]c_{1} \approx 157.532[/tex], [tex]c_{2}\approx 36.676[/tex]

Step-by-step explanation:

The Law of Sines states that:

[tex]\frac{a}{\sin A} = \frac{b}{\sin B}=\frac{c}{\sin C}[/tex]

Where:

[tex]a[/tex], [tex]b[/tex], [tex]c[/tex] - Side lengths, dimensionless.

[tex]A[/tex], [tex]B[/tex], [tex]C[/tex] - Angles opposite to respective sides, dimensionless.

Given that [tex]a = 71[/tex], [tex]b = 104[/tex], [tex]\angle A = 21^{\circ}[/tex], the sine of angle B is:

[tex]\sin B = \frac{b}{a}\cdot \sin A[/tex]

[tex]\sin B = \frac{104}{71}\cdot \sin 21^{\circ}[/tex]

[tex]\sin B = 0.525[/tex]

Sine is positive between 0º and 180º, so there are two possible solutions:

[tex]\angle B_{1} \approx 31.668^{\circ}[/tex]

[tex]\angle B_{2} \approx 148.332^{\circ}[/tex]

The remaining angle is obtained from the principle that sum of internal triangles equals to 180 degrees: ([tex]\angle A = 21^{\circ}[/tex], [tex]\angle B_{1} \approx 31.668^{\circ}[/tex], [tex]\angle B_{2} \approx 148.332^{\circ}[/tex])

[tex]\angle C_{1} = 180^{\circ}-\angle A - \angle B_{1}[/tex]

[tex]\angle C_{1} = 180^{\circ}-21^{\circ}-31.668^{\circ}[/tex]

[tex]\angle C_{1} \approx 127.332^{\circ}[/tex]

[tex]\angle C_{2} = 180^{\circ}-\angle A - \angle B_{2}[/tex]

[tex]\angle C_{2} = 180^{\circ}-21^{\circ}-148.332^{\circ}[/tex]

[tex]\angle C_{2} \approx 10.668^{\circ}[/tex]

Lastly, the remaining side of the triangle is found by means of the Law of Sine: ([tex]a = 71[/tex], [tex]\angle A = 21^{\circ}[/tex], [tex]\angle C_{1} \approx 127.332^{\circ}[/tex], [tex]\angle C_{2} \approx 10.668^{\circ}[/tex])

[tex]c_{1} = a\cdot \left(\frac{\sin C_{1}}{\sin A} \right)[/tex]

[tex]c_{1} = 71\cdot \left(\frac{\sin 127.332^{\circ}}{\sin 21^{\circ}} \right)[/tex]

[tex]c_{1} \approx 157.532[/tex]

[tex]c_{2} = a\cdot \left(\frac{\sin C_{2}}{\sin A} \right)[/tex]

[tex]c_{2}= 71\cdot \left(\frac{\sin 10.668^{\circ}}{\sin 21^{\circ}} \right)[/tex]

[tex]c_{2}\approx 36.676[/tex]

The answer are presented below:

[tex]\angle B_{1} \approx 31.668^{\circ}[/tex], [tex]\angle B_{2} \approx 148.332^{\circ}[/tex]

[tex]\angle C_{1} \approx 127.332^{\circ}[/tex], [tex]\angle C_{2} \approx 10.668^{\circ}[/tex]

[tex]c_{1} \approx 157.532[/tex], [tex]c_{2}\approx 36.676[/tex]

ACCESS MORE