Question

Solve triangle ABC which has angleA=250.251,angleB=600.511anda=3.82.Find c.

A 3-6cm
B 7.0cm
C 7.4cm
D 8.8cm

Answer 1

Answer on Question #51807 – Math – Trigonometry

Solve triangle ABC which has angle A=250.251, angle B=600.511 and a=3.82. Find c.

A 3-6cm

B 7.0cm

C 7.4cm

D 8.8cm

Solution

The given values of angles $\hat{\mathbf{A}}$ and $\mathbf{B}$ are unrealistic.

Triangle with such values for angles does not exist, because the sum of interior angles of triangle is equal to $180{}^{\circ}$.

You're asked to find c, so let's get angle C from triangle's property:

$$C = 180{}^{\circ} - (A + B).$$

By the law of sines, then use

$$\frac{\sin(A)}{a} = \frac{\sin(C)}{c},$$

which gives the expression for $c$:

$$c = \frac{a \cdot \sin(C)}{\sin(A)} = \frac{a \cdot \sin(180{}^{\circ} - (A + B))}{\sin(A)} = \frac{a \cdot \sin((A + B))}{\sin(A)} = \frac{a \cdot (\sin(A)\cos(B) + \cos(A)\sin(B))}{\sin(A)},$$

because $\sin(180{}^{\circ} - \alpha) = \sin(\alpha)$, $\sin(\alpha + \beta) = \sin(\alpha)\cos(\beta) + \cos(\alpha)\sin(\beta)$.

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