Question

math 20-2 trigonometry assignment
name:
10 a triangular playground has sides of lengths 475 feet, 596 feet, and 401 feet. what are the measures of the angles between the sides, to the nearest tenth of a degree?
11 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.
12 a clock has a minute hand that is 20 cm long and an hour hand that is 12 cm long. determine the direct straight line distance between the hands at 5:00 pm.

Explanation:

Problem 10

Step1: Assign sides to triangle

Let $a=475$ ft, $b=595$ ft, $c=401$ ft; angles opposite: $A,B,C$.

Step2: Find angle A via Law of Cosines

Law of Cosines: $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$
$\cos A = \frac{595^2 + 401^2 - 475^2}{2(595)(401)} = \frac{354025 + 160801 - 225625}{477190} = \frac{289201}{477190} \approx 0.606$
$A = \arccos(0.606) \approx 52.7^\circ$

Step3: Find angle B via Law of Cosines

$\cos B = \frac{a^2 + c^2 - b^2}{2ac}$
$\cos B = \frac{475^2 + 401^2 - 595^2}{2(475)(401)} = \frac{225625 + 160801 - 354025}{380950} = \frac{32401}{380950} \approx 0.085$
$B = \arccos(0.085) \approx 85.1^\circ$

Step4: Find angle C via angle sum

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

Step1: Find angle at B in $\triangle BCD$

Angle at $B$: $180^\circ - 50^\circ = 130^\circ$

Step2: Find angle at D in $\triangle BCD$

Angle sum: $\angle BDC = 180^\circ - 130^\circ - 63^\circ = -13^\circ$ (correction: use $\triangle ABD$ first. Let the base point be $C$, so $\triangle ABC$ and $\triangle ADC$ are right triangles. First find $AB$ using $\triangle BCD$:
Law of Sines on $\triangle BCD$: $\frac{BD}{\sin(180-50-63)} = \frac{BC}{\sin63^\circ}$
$\frac{200}{\sin67^\circ} = \frac{BC}{\sin63^\circ}$
$BC = \frac{200\sin63^\circ}{\sin67^\circ} \approx \frac{200(0.8910)}{0.9205} \approx 193.6$ m

Step3: Calculate height h

In right $\triangle ABC$: $\sin(95^\circ) = \frac{h}{AB}$ (correction: $\angle ABC=95^\circ$ is the angle between $AB$ and $BC$, so $h = AB\sin(95^\circ)$? No, right triangle: $h = BC\tan(95^\circ)$? No, $AB$ is the rope, so $h = AB\sin(\theta)$ where $\theta$ is the angle between rope and vertical. Correct: In $\triangle ABC$, right-angled at $C$, $\angle BAC = 90^\circ - 32^\circ = 58^\circ$? No, use Law of Cosines on $\triangle ABD$ first:
$AB=95$ m, $BD=200$ m, $\angle ABD=50^\circ$
Law of Cosines: $AD^2 = AB^2 + BD^2 - 2(AB)(BD)\sin(50^\circ)$
$AD^2 = 95^2 + 200^2 - 2(95)(200)\cos50^\circ = 9025 + 40000 - 38000(0.6428) = 49025 - 24426.4 = 24598.6$
$AD \approx 156.8$ m
Now use Law of Sines on $\triangle ABD$: $\frac{\sin\angle BAD}{BD} = \frac{\sin50^\circ}{AD}$
$\sin\angle BAD = \frac{200\sin50^\circ}{156.8} \approx \frac{200(0.7660)}{156.8} \approx 0.975$
$\angle BAD \approx 77.2^\circ$
Angle between $AB$ and vertical: $77.2^\circ - 32^\circ = 45.2^\circ$
$h = AB\sin(45.2^\circ) = 95(0.710) \approx 67.5$ m
Correction: Simpler: In $\triangle BCD$, $\angle CBD=50^\circ$, $\angle CDB=63^\circ$, so $\angle BCD=67^\circ$
Law of Sines: $\frac{BC}{\sin63^\circ} = \frac{200}{\sin67^\circ}$
$BC = \frac{200\sin63^\circ}{\sin67^\circ} \approx 193.6$ m
In right $\triangle ABC$, $\angle ABC=95^\circ$, so $h = BC\tan(95^\circ)$? No, $\angle ABC$ is the angle between $AB$ and $BC$, so $\sin(95^\circ) = \frac{h}{AB}$ → $h = 95\sin(95^\circ) \approx 94.6$ m (wrong, because $AB$ is not the hypotenuse). Correct: The vertical line from $A$ is $h$, so $\angle ABh = 95^\circ - 90^\circ = 5^\circ$? No, use the two right triangles:
Let $x$ be the horizontal distance from $B$ to the base of $h$, $y$ be the horizontal distance from $D$ to the base of $h$.
$x = h\cot(95^\circ)$, $y = h\cot(32^\circ + 63^\circ) = h\cot(95^\circ)$? No, $\angle ADB=63^\circ$, so $y = h\cot(63^\circ)$
$x + y = 200$ → $h(\cot(95^\circ) + \cot(63^\circ)) = 200$
$\cot(95^\circ) = \tan(-5^\circ) \approx -0.0875$, $\cot(63^\circ) \approx 0.5095$
$h(0.5095 - 0.0875) = 200$ → $h(0.422) = 200$ → $h \approx 473.9$ m (wrong, sign error: $\angle ABh = 90^\circ - 95^\circ$ is negative, so $|x| + y = 200$ → $h(|\cot(95^\circ)| + \cot(63^\circ)) = 200$
$h(0.0875 + 0.5095) = 200$ → $h(0.597) = 200$ → $h \approx 335.0$ m
Final correct method: Use Law of Cosines on $\triangle ABD$:
$AB=95$, $BD=200$, $\angle ABD=50^\circ$, so $AD^2=95^2+200^2-2*95*200*\cos50^\circ=9025+40000-38000*0.6428=49025-24426.4=24598.6$, $AD≈156.8$
Now in $\triangle ABD$, use Law of Sines to find $\angle BAD$:
$\frac{\sin\angle BAD}{200}=\frac{\sin50^\circ}{156.8}$ → $\sin\angle BAD≈(200*0.7660)/156.8≈0.975$, $\angle BAD≈77.2^\circ$
The angle between $AB$ and the vertical is $77.2^\circ - 32^\circ=45.2^\circ$
$h=AB*\cos(45.2^\circ)=95*0.704≈66.9$ m

Step1: Find angle between clock hands

At 5:00, the hour hand is at 5, minute at 12. Total clock angle: $360^\circ$, 12 intervals, so each interval is $30^\circ$.
Angle between hands: $5*30^\circ=150^\circ$

Step2: Use Law of Cosines for distance

Let $d$ = distance, $a=20$ cm, $b=12$ cm, $\theta=150^\circ$
Law of Cosines: $d^2 = a^2 + b^2 - 2ab\cos\theta$
$d^2 = 20^2 + 12^2 - 2(20)(12)\cos150^\circ$
$\cos150^\circ=-\frac{\sqrt{3}}{2}≈-0.8660$
$d^2 = 400 + 144 - 480(-0.8660) = 544 + 415.68 = 959.68$
$d = \sqrt{959.68}≈30.98$ cm

Answer:

The angles are approximately $52.7^\circ$, $85.1^\circ$, and $42.2^\circ$

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Problem 11