Question

for questions 5 - 6, solve for the remaining angles. (3 points each)
5.
triangle abc with side lengths: bc = 270, ac = 442.85, ab = 255
6.
isosceles triangle abc with ab = 15, ac = 15, bc = 19

Explanation:

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Question 5 Steps:

Step1: Find $\angle A$ via Law of Cosines

$\cos A = \frac{255^2 + 442.85^2 - 270^2}{2 \times 255 \times 442.85}$
$\cos A \approx \frac{65025 + 196116.12 - 72900}{225853.5} \approx \frac{188241.12}{225853.5} \approx 0.8335$
$\angle A \approx \arccos(0.8335) \approx 35.9^\circ$

Step2: Find $\angle C$ via Law of Cosines

$\cos C = \frac{270^2 + 442.85^2 - 255^2}{2 \times 270 \times 442.85}$
$\cos C \approx \frac{72900 + 196116.12 - 65025}{239139} \approx \frac{203991.12}{239139} \approx 0.8530$
$\angle C \approx \arccos(0.8530) \approx 34.1^\circ$

Step3: Calculate $\angle B$

$\angle B = 180^\circ - 35.9^\circ - 34.1^\circ = 110^\circ$
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Question 6 Steps:

Step1: Find $\angle A$ via Law of Cosines

$\cos A = \frac{15^2 + 15^2 - 19^2}{2 \times 15 \times 15}$
$\cos A = \frac{225 + 225 - 361}{450} = \frac{89}{450} \approx 0.1978$
$\angle A \approx \arccos(0.1978) \approx 80.2^\circ$

Step2: Calculate $\angle B$ and $\angle C$

Since $AB=AC$, $\angle B = \angle C$
$\angle B = \angle C = \frac{180^\circ - 80.2^\circ}{2} \approx 49.9^\circ$

Answer:

Question 5: $\angle A \approx 35.9^\circ$, $\angle C \approx 34.1^\circ$, $\angle B = 110^\circ$
Question 6: $\angle A \approx 80.2^\circ$, $\angle B = \angle C \approx 49.9^\circ$