Question

triangle abc is shown below with all three side lengths known.

image of triangle abc with a to c = 13 cm, a to b = 9 cm, b to c = 10.2 cm

determine the missing angle measures using the law of cosines, law of sines, and other properties of triangles.
a. ( mangle a = ) blank degrees preview
b. ( mangle b = ) blank degrees preview
c. ( mangle c = ) blank degrees preview
this question allows you to use trig functions that evaluate for angle measures in degrees. to remind yourself how to evaluate trig functions for angle...
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Explanation:

Step1: Find \( m\angle B \) using Law of Cosines

The Law of Cosines states \( AC^2 = AB^2 + BC^2 - 2 \cdot AB \cdot BC \cdot \cos(\angle B) \). Here, \( AC = 13 \), \( AB = 9 \), \( BC = 10.2 \).

$$ 13^2 = 9^2 + 10.2^2 - 2 \cdot 9 \cdot 10.2 \cdot \cos(\angle B) $$

$$ 169 = 81 + 104.04 - 183.6 \cdot \cos(\angle B) $$

$$ 169 = 185.04 - 183.6 \cdot \cos(\angle B) $$

$$ -16.04 = -183.6 \cdot \cos(\angle B) $$

$$ \cos(\angle B) = \frac{16.04}{183.6} \approx 0.0874 $$

$$ m\angle B \approx \arccos(0.0874) \approx 85^\circ $$

Step2: Find \( m\angle A \) using Law of Sines

Law of Sines: \( \frac{\sin(\angle A)}{BC} = \frac{\sin(\angle B)}{AC} \)

$$ \frac{\sin(\angle A)}{10.2} = \frac{\sin(85^\circ)}{13} $$

$$ \sin(\angle A) = \frac{10.2 \cdot \sin(85^\circ)}{13} \approx \frac{10.2 \cdot 0.9962}{13} \approx 0.783 $$

$$ m\angle A \approx \arcsin(0.783) \approx 51.5^\circ $$

Step3: Find \( m\angle C \) using triangle angle sum

Triangle angle sum: \( m\angle A + m\angle B + m\angle C = 180^\circ \)

$$ m\angle C = 180^\circ - 51.5^\circ - 85^\circ \approx 43.5^\circ $$

(Note: Due to rounding, values may vary slightly. More precise calculations:
For \( \angle B \):

$$ \cos B = \frac{9^2 + 10.2^2 - 13^2}{2 \cdot 9 \cdot 10.2} = \frac{81 + 104.04 - 169}{183.6} = \frac{16.04}{183.6} \approx 0.08736 $$

\( B \approx 85.0^\circ \) (using calculator for \( \arccos(0.08736) \))

For \( \angle A \):

$$ \sin A = \frac{10.2 \cdot \sin 85.0^\circ}{13} \approx \frac{10.2 \cdot 0.9961947}{13} \approx 0.783 $$

\( A \approx 51.5^\circ \) (or more precisely, using Law of Cosines for \( A \):

$$ \cos A = \frac{9^2 + 13^2 - 10.2^2}{2 \cdot 9 \cdot 13} = \frac{81 + 169 - 104.04}{234} = \frac{145.96}{234} \approx 0.6237 $$

\( A \approx \arccos(0.6237) \approx 51.5^\circ \)

Then \( C = 180 - 51.5 - 85 = 43.5^\circ \))

Answer:

a. \( m\angle A \approx \boldsymbol{51.5} \) degrees
b. \( m\angle B \approx \boldsymbol{85} \) degrees
c. \( m\angle C \approx \boldsymbol{43.5} \) degrees

(Note: Depending on rounding during steps, values might be slightly different. For example, more precise calculation for \( \angle A \) using Law of Cosines:
\( \cos A = \frac{AB^2 + AC^2 - BC^2}{2 \cdot AB \cdot AC} = \frac{9^2 + 13^2 - 10.2^2}{2 \cdot 9 \cdot 13} = \frac{81 + 169 - 104.04}{234} = \frac{145.96}{234} \approx 0.6237 \), so \( A = \arccos(0.6237) \approx 51.5^\circ \).
For \( \angle B \), \( \cos B = \frac{9^2 + 10.2^2 - 13^2}{2 \cdot 9 \cdot 10.2} \approx 0.0874 \), \( B \approx 85^\circ \).
For \( \angle C \), \( 180 - 51.5 - 85 = 43.5^\circ \).)