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SCIE 4101 Math Review Packet #6 Solutions October 30, 2014

Posted by Ms. Miller in UTA.
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1. \tan 24\textdegree\approx 0.4452 2. \sin 16\textdegree\approx 0.2756
3. \cos 81\textdegree\approx 0.1564 4. \tan^{-1}2.4751\approx 68\textdegree
5. \sin{-1}0.3746\approx 22\textdegree 6. \cos^{-1}0.8290\approx 34\textdegree
7a. \sin M=\frac{48}{50}=\frac{24}{25} b. \cos M=\frac{14}{50}=\frac{7}{25}
c. \tan M=\frac{48}{14}=\frac{24}{7} d. m\angle M=\cos^{-1}\frac{7}{25}\approx 74\textdegree
e. Either m\angle V=90\textdegree-74\textdegree = 16\textdegree
or m\angle V=\sin^{-1}\frac{7}{25}\approx 16\textdegree
f. \cos V=\frac{48}{50}=\frac{24}{25}
8.

  • missing side: \sqrt{5^2+8^2}\approx 9.4
  • angle opp. 5: \tan^{-1}\frac{5}{8}\approx 32\textdegree
  • angle opp. 8: 90-32=58\textdegree
9.

  • missing side: \sqrt{8^2-5^2}\approx 6.2
  • angle opp. 5: \sin^{-1}\frac{5}{8}\approx 39\textdegree
  • other angle: 90-39=51\textdegree
10.

  • missing angle: 90-54=36\textdegree
  • side opp. 54°: \tan 54\textdegree=\frac{x}{10}
    x\approx 13.8
  • hypotenuse: \cos 54\textdegree=\frac{10}{y}
    y\approx 17.0
11.

  • missing side: \sqrt{22^2-12^2}\approx 18.4
  • angle opp. 12: \sin^{-1}\frac{12}{22}\approx 33\textdegree
  • other angle: 90-33=57\textdegree
12.

  • missing angle: 90-39=51\textdegree
  • missing leg: \tan 39\textdegree=\frac{10}{x}
    x\approx 12.3
  • hypotenuse: \sin 39\textdegree=\frac{10}{y}
    x\approx 15.9
13.

  • missing side: \sqrt{(3\sqrt{5})^2-3^2}=\sqrt{45-9}=6
  • angle opp. 3: \sin^{-1}\frac{3}{3\sqrt{5}}\approx 27\textdegree
  • other angle: 90-27=63\textdegree
14.

  • missing side: \sqrt{14^2-5^2}\approx 13.1
  • angle opp. 5: \sin^{-1}\frac{5}{14}\approx 21\textdegree
  • other angle: 90-21=69\textdegree
15.

  • hypotenuse: \sqrt{6^2+13^2}\approx 14.3
  • angle opp. 6: \tan^{-1}\frac{6}{13}\approx 25\textdegree
  • angle opp. 13: 90-25=65\textdegree
16.

  • hypotenuse: \sqrt{7^2+11^2}\approx 13.0
  • angle opp. 7: \tan^{-1}\frac{7}{11}\approx 32\textdegree
  • angle opp. 11: 90-32=58\textdegree
17.

  • missing angle: 90-56=34\textdegree
  • missing leg: \tan 56\textdegree=\frac{x}{10}
    x\approx 14.8
  • hypotenuse: \cos 56\textdegree=\frac{10}{y}
    y\approx 17.9
18. \frac{VS}{ES}=\sin E
\frac{VS}{3\times 10^8}=\sin 21\textdegree
VS\approx 1.08\times 10^8\ \mathrm{km}
19.
To solve this, we will use Law of Cosines: c^2=a^2+b^2-2ab\cos C
c^2=(254.5)^2+(251.5)^2-2(254.5)(251.5)\cos 103\textdegree
c^2=64770.25+63252.25-128013.5\cos 103\textdegree
c^2\approx 156819
c\approx 396\ \mathrm{mi}
20. We can find the other angles by subtracting 39° from 180° and dividing by 2: \frac{180-39}{2}=70.5\textdegree.
To find the length of each side, we will use Law of Sines:
\frac {x}{\sin 70.5\textdegree}=\frac{425}{\sin 39\textdegree}
x=\sin 70.5\textdegree(\frac{425}{\sin 39\textdegree})
x\approx 637\ \mathrm{ft}
21.This problem is a two-step process. First we find x by using Law of Sines, then we find y by using sine:
We know the angle next to the 70° angle is 110°. That means the angle opposite the 210 foot side is 10° (or we could figure that moving from 60° to 70° was 10°).
\frac{x}{\sin 60\textdegree}=\frac{210}{\sin 10\textdegree}
x=\sin 60\textdegree(\frac{210}{\sin 10\textdegree})
x\approx 1047.3\ \mathrm{ft}
\sin 70\textdegree=\frac{y}{1047.3}
y\approx 984\ \mathrm{ft}
22. Use Law of Cosines:
c^2=82^2+97^2-2(82)(97)\cos 125\textdegree
c^2=6724+9409-15908\cos 125\textdegree
c^2\approx 25257.5
c\approx 159\ \mathrm{feet}
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