## SCIE 4101 Math Review Packet #6 SolutionsOctober 30, 2014

Posted by Ms. Miller in UTA.
 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: $c^2=82^2+97^2-2(82)(97)\cos 125\textdegree$ $c^2=6724+9409-15908\cos 125\textdegree$ $c^2\approx 25257.5$ $c\approx 159\ \mathrm{feet}$

## Obj. 14 Equations of LinesOctober 28, 2014

Posted by Ms. Miller in Geometry.

Practice: Khan Academy Slope-Intercept Form

## SCIE 4101 Review Packet #6October 28, 2014

Posted by Ms. Miller in UTA.

Review Packet 6 – Trigonometry (2014)

Review Packet 6 – Trigonometry (2014) Notes

## Obj. 13 Perpendicular LinesOctober 26, 2014

Posted by Ms. Miller in Geometry.

Inequality Practice: IXL.com Multi-step Inequalities

Perpendicular Practice: Khan Academy Recognizing Parallel and Perpendicular Lines

## Obj. 12 Proving Lines ParallelOctober 23, 2014

Posted by Ms. Miller in Geometry.

## SCIE 4101 Review Packet #5 – Algebra II #2October 22, 2014

Posted by Ms. Miller in UTA.

Review Packet 5 – Algebra II (2) (2014) Notes

Review Packet 5 – Algebra II (2) (2014)

Review Packet 5 Solutions (2014)

## Obj. 11 Parallel Lines and TransversalsOctober 16, 2014

Posted by Ms. Miller in Geometry.

Practice: IXL.com Transversals of Parallel Lines

## SCIE 4101 Review Packet #4 SolutionsOctober 14, 2014

Posted by Ms. Miller in UTA.

Review Packet 4 Solutions

Notice: You may not use a calculator on the quiz this Monday. All of the problems will have integer solutions, and I do not consider any of the multiplication or division too difficult for potential math or science teachers to do.

## Chapter 2 Test Review and SolutionsOctober 13, 2014

Posted by Ms. Miller in Geometry.

In case you lost your copy: PAP Chapter 2 Test Review (A-day, you will get this during class on Tuesday.)

Chapter 2 Test Review Solutions

## SCIE 4101 Review Packet #4October 13, 2014

Posted by Ms. Miller in UTA.