Question

In ∆ABC, if the lengths of a, b, and c are 22.5 centimeters, 18 centimeters, and 13.6 centimeters, respectively, what are m < B and m 12 years ago

Answer 1

m B = 89.68°, and m C = 37.19°

m B = 53.12°, and m C = 89.68°

m B = 89.68°, and m C = 53.13°

m B = 37.20°, and m C = 53.12°

m B = 53.13°, and m C = 37.19°

Answer 2

Use the law of cosines to find the angles. \[a^2+b^2 -2 a b \cos C = c^2\]

Answer 3

so b

Answer 4

Mmm, no. Look, angle C is clearly the smallest angle in the drawing, right? B is the next largest, and A is by far the largest. That alone should guide you to the last answer.

However, if I actually had to calculate them, here's how I would do it.

\[a^2 + b^2 - 2ab \cos C = c^2\]Remember, C is the angle opposite side c.
\[\frac{c^2-a^2-b^2}{-2ab} = \cos C\]
\[\cos C = \frac{13.6^2-22.5^2-18^2}{-2(22.5)(18)} \approx 0.797\]
\[C = \cos^{-1} 0.797 =0.649\text{ rad} = 37.19^\circ\]

Now to find B:
\[a^2+c^2 -2ac \cos B = b^2\]\[\cos B = \frac{b^2-a^2-c^2}{-2ac}\]\[B = \cos^{-1} \frac{b^2-a^2-c^2}{-2ac} = \cos^{-1} \frac{18^2-22.5^2-13.6^2}{-2(22.5)(13.6)} \approx 0.927 \text{ rad} = 53.13^\circ \]

and finally A:
\[b^2+c^2-2bc \cos A = a^2\]\[A = 89.69^\circ\]

Answer 5

sorry, rounded incorrectly, \[A = 89.68^\circ\]
It's late, time to sleep :-)