Question

name:

1. a triangular playground has sides of lengths 475 feet, 595 feet, and 401 feet. what are the measures of the angles between the sides, to the nearest tenth of a degree?
2. the hot air balloon pictured on the right is anchored to the ground by ropes ab and ad. determine the height h, of the balloon.

(image of a hot air balloon anchored with ropes ab and ad, with triangle details: ab=95 m, angle at b is 50°, bd=200 m, angle at a between h and ad is 32°, angle at d is 63°)

Explanation:

Problem 10

Step1: Assign sides to angles

Let $a=595$ ft, $b=475$ ft, $c=401$ ft. Angles opposite: $\angle A$, $\angle B$, $\angle C$.

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

$\cos A = \frac{b^2 + c^2 - a^2}{2bc}$
$\cos A = \frac{475^2 + 401^2 - 595^2}{2(475)(401)}$
$\cos A = \frac{225625 + 160801 - 354025}{380950} = \frac{32401}{380950} \approx 0.08505$
$\angle A = \arccos(0.08505) \approx 85.1^\circ$

Step3: Find $\angle B$ via Law of Cosines

$\cos B = \frac{a^2 + c^2 - b^2}{2ac}$
$\cos B = \frac{595^2 + 401^2 - 475^2}{2(595)(401)}$
$\cos B = \frac{354025 + 160801 - 225625}{477190} = \frac{289201}{477190} \approx 0.6060$
$\angle B = \arccos(0.6060) \approx 52.7^\circ$

Step4: Find $\angle C$ via angle sum

$\angle C = 180^\circ - \angle A - \angle B$
$\angle C = 180^\circ - 85.1^\circ - 52.7^\circ = 42.2^\circ$

Step1: Find $\angle ABD$ in $\triangle BCD$

$\angle BCD = 180^\circ - 50^\circ - 63^\circ = 67^\circ$

Step2: Find length $BD$ via Law of Sines

$\frac{BD}{\sin 67^\circ} = \frac{95}{\sin 63^\circ}$
$BD = \frac{95 \sin 67^\circ}{\sin 63^\circ} \approx \frac{95(0.9205)}{0.8910} \approx 98.7$ m

Step3: Find $h$ using right triangle

$h = BD \sin 50^\circ$
$h \approx 98.7(0.7660) \approx 75.6$ m

Answer:

The angles are approximately $\boldsymbol{85.1^\circ}$, $\boldsymbol{52.7^\circ}$, and $\boldsymbol{42.2^\circ}$.

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Problem 11