Question

Find the missing parts of the triangle. (Find angles to the nearest hundredth of a degree.)

a = 7.5

b = 13.3

c = 16.0

Answer 1

Answer:

$A \approx 27.75^{\circ}$, $B \approx 55.67^{\circ}$, $C \approx 96.60^{\circ}$

Step-by-step explanation:

The first two angle can be determined by the Law of the Cosine:

$\cos A = \frac{7.5^{2}-13.3^{2}-16^{2}}{-2\cdot (13.3)\cdot (16)}$

$\cos A = 0.885$

$A = \cos^{-1} 0.885$

$A \approx 27.75^{\circ}$

$\cos B =\frac{13.3^{2}-7.5^{2}-16^{2}}{-2\cdot (7.5)\cdot (16)}$

$\cos B = 0.564$

$B = \cos^{-1} 0.564$

$B \approx 55.67^{\circ}$

$\cos C = \frac{16^{2}-7.5^{2}-13.3^{2}}{-2\cdot (7.5)\cdot (13.3)}$

$\cos C = -0.115$

$C = \cos^{-1} (-0.115)$

$C \approx 96.60^{\circ}$

The sum of internal angles of a triangle must be equal to 180°. Then:

$A + B + C = 27.75^{\circ} + 55.67^{\circ} + 96.60^{\circ}$

$A + B + C = 180.02^{\circ}$

Which satisfies the condition described.