Answers for the Third Examination

1. a) $x=2$                          b) $x=-\frac{1}{4}$                         c) $x = 5$

d) $x=6$                         e) $x=40$                          f) $x=2$

g) $x=-2$ and $x = 5$         h) $x=\pm 3$                          i) $x=2$ and $x=3$

j) $x=1$ and $x = 5$         k) $x=2$ and $x = 5$          l) $x=3$

m) $x=2$                         n) $x=e+2$                  o) $x=3+e^{2}$

p) $x=\frac{1}{e}$

2. a) $x=4.80$                          b) $x=.85$                          c) $x=1.39$

d) $x=\frac{1}{2}e^{10}-\frac{1}{2}=11013.$

3. a) $\frac{\left( 5^{2}\right) ^{3}\cdot 10^{-2}}{\left( 5\sqrt{2}%% \right) ^{4}}=\frac{1}{16}$                  b) $\frac{\left(3^{3}\right) ^{2}\cdot 6^{-4}}{\left( 2\sqrt{3}\right) ^{-4}}=81$

c) $(2^{\frac{3x}{2}})^{\frac{2}{x}}-\left( \frac{1}{2}%% \right) ^{x}\cdot \frac{4^{x}}{2^{x}}= 7$

d) $\ \log _{a}\left( 3a\right) ^{4}-2\log _{a}\left( 9\right) +\log _{a}\left( \frac{1}{a^{2}}\right) =2$

e) $\log_{a}\left( 2a\right) ^{3}-2\log _{a}\left( 4\right) + \log _{a}\left( \frac{2}{a}\right) =2$

f) $\ln \left( x^{\frac{5}{2}}\right) -2\ln \left( x^{3}\right) + \frac{3}{2}\ln \left( x^{4}\right) = \frac{5}{2}\ln x$

g) $\ \ln (e^{3x})-3e^{3\ln \left( \sqrt[3]{x}\right) }=0$         h) $\log (10^{3x})-2e^{2\ln (\sqrt{x)}}=x$

i) $\log_{83}(83)=1$                  j) $\log _{3}\left( 243\right) 5$

k) $\log _{10}\left( .001\right) =-3$                  l) $\log _{a}(a^{5})= 5$

m) $\log _{8}\left( \frac{1}{16}\right) = -\frac{4}{3}$         n) $\log _{27}\left( \frac{1}{3}\right) = -\frac{1}{3}$

o) $\ln \left( \sqrt{e}\right)=\frac{1}{2}$                         p) $\ln (e^{3})=3$

q) $\ln \left( \frac{1}{e}\right) = -1$                          r) $\ 7^{2\log _{7}7}= 49$

s) $\ \sqrt{e^{\frac{1}{2}\ln \left( 9\right) }}=\sqrt{3}$                  t) $\left( \frac{e^{x}+e^{-x}}{2}\right) ^{2}- \left( \frac{e^{x}-e^{-x}}{2}\right) ^{2}=1$

4. Given that $\ln \left( 2\right) \approx .693\,$, $\ln \left( 3\right) \approx 1.099$ and $\ln \left( 5\right) \approx 1.609$, find each of the following to two decimal places (without using a calculator):

a) $\ln \left( 12\right) =2.49$                          b) $\ln \left( 30\right) =3.40$

c) $\ln \left( \frac{3}{2}\right) =.41$                 d) $\ln \left( \sqrt[3]{2}\right) =.23$

e) $\ln \left( e^{12}\right) =12$                          f) $e^{\ln \left( 5\right) }=5$

g) $\ln \left( 5^{2}\right) =3.22$                          h) $\left( \ln \left( 5\right) \right) ^{2}=2.59$

i) $\log _{3}\left( 5\right)=1.46$

5. Given that $\ln \left( a\right) \approx .31$ and $\ln \left( b\right) \approx .69$ find each of the following to two decimal places :

a) $\ln \left( a^{2}b\right) = 1.31$                  b) $\ln \left( \frac{a}{b^{2}}\right) = -1.07$                  c) $\ln \left( \sqrt[3]{b}\right) =.23$

d) $\ e^{\ln (\ln \left( ab\right) )}=1.00$                  e) $\log_{a}\left( b\right) =2.23$

6. Given that $\ln \left( 2\right) =u$ and that $\ln \left( 7\right) =v$, express each of the following in terms of $u$ and $v$ (Do not use a calculator):

a) $\ln \left( 28\right) =2u+v$          b) $\ln \left( \frac{7}{8}\right) =v-3u$          c) $\ln \left( \frac{\sqrt{8}}{49}\right) =\frac{3}{2}u-2v$

d) $\ln \left( 7^{\ln \left( 2\right) }\right) =uv$          e) $e^{2u}=4$                  f) $\log _{2}\left( 7\right) =\frac{v}{u}$

7. a) $\ g\left( x\right) =\left( \frac{1}{e}\right) ^{x}$

asymptotes: $y=0$; $\qquad y$-intercept: $\left( 0,1\right)$

b) $p\left( x\right) =3^{x}$

asymptotes: $y=0$; $\qquad y$-intercept: $\left( 0,1\right)$

c) $\ f\left( x\right) =2+e^{1-x}$

asymptotes: $y=2$; $\qquad y$-intercept: $\left( 0,e+2\right)$

d) $h(x)=\left( \frac{3}{4}\right) ^{-x}$

asymptotes: $y=0$; $\qquad y$-intercept: $\left( 0,1\right)$

e) $k\left( x\right) =\ln \left( 2x-3\right)$

asymptotes: $x=\frac{3}{2}$; $\qquad x$-intercept: $\left( 2,0\right)$

f) $m\left( x\right) =\ln \left( 1-x\right)$

asymptotes: $x=1$; $x$-intercept and $y$-intercept: $\left( 0,0\right)$

g) $r\left( x\right) =\ln \left( -x\right)$

asymptotes: $x=0$; $x$-intercept: $\left( 0,-1\right)$

h) $\ s\left( x\right) =1+\ln \left( x-3\right)$

asymptotes: $x=3$; $x$-intercept: $\left( 3+\frac{1}{e},0\right)$

i) $\ t\left( x\right) =\ln \left( \frac{1}{e}\right) ^{x}$

no asymptotes;         $x$- and $y$- intercepts: $\left( 0,0\right)$

j) $q\left( x\right) =e^{\ln \left( \left\vert x\right\vert \right) }$

no asymptotes;         $x$- and $y$- intercepts: $\left( 0,0\right)$

k) $j\left( x\right) =3^{\log _{3}(x+1)}$

no asymptotes;         $x$-intercept: $\left( -1,0\right)$;                  $y$- intercept: $\left( 0,1\right)$

l) $u(x)=e^{(x^{2})}$

no asymptotes;         $x$- and $y$- intercepts: $\left( 0,0\right)$

m) $v(x)=\ln \left( x^{2}\right)$

asymptote: $x=0$;         $x$-intercepts: $\left( -1,0\right)$ and $\left( 1,0\right)$

n) $n\left( x\right) =(\ln \left( x\right) )^{2}$

asymptote: $x=0$;         $x$-intercept: $\left( 1,0\right)$

o) $w\left( x\right) =\ln \left( e^{x^{2}-8x+15}\right)$

no asymptotes;         $x$-intercepts: $\left( 3,0\right)$ and $\left( 5,0\right)$

8. a) To the nearest tenth of a percent, at what rate of interest compounded continuously will $1000 grow to be worth $1200 after 4 years?

$1000 \ast e^{4r}=1200$

$r\approx .04558\approx 4.6$

b) If $5000 is invested at an interest rate of 10% compounded annually and if the interest rate decreases to 8% compounded annually after two years, how much will the investment be worth at the end of 5 years?

$5000\ast (1+.100)^{2}\left( 1+.0800\right) ^{3}$

$\$7621.26$

c) If $5000 is invested at an interest rate of 6% compounded continuously and if the interest rate doubles after six years, how much will the investment be worth at the end of ten years?

$5000\ast e^{(.06)\ast 6}\ast e^{\left( .12\right) \ast 4}$

$\$11581.83$

d) If $2000 is invested at 8% interest, find the value of the investment after 2 years (to the nearest cent) if

        i) interest is compounded semiannually

$2000\ast (1+.04)^{4}$

$\$2339.72$

        ii) interest is compounded continuously

$2000\ast e^{\left( .08\right) \ast 2}$

$\$2347.02$

e) The bacteria in a petri dish are dying at a constant rate due to an antibiotic which has been introduced. If half of the bacteria remaining in the dish die every three hours, to the nearest tenth of an hour, how long will it take until only 10% of the original number of bacteria remain?

$.50N=N\ast e^{3\ast k}$

$k\approx -.231\,05$

$.10N\approx N\ast e^{\left( -.23105\right) \ast t}$

$t\approx 10.0$ hours

f) The number of bacteria in a petri dish doubles after two days. To the nearest tenth of a day, how long does is take until the number of bacteria is ten times to original amount.

$2\ast N=N\ast e^{2k}$

$k=\frac{1}{2}\ln 2$

$10\ast N=N\ast e^{\left( \frac{1}{2}\ln 2\right) \ast t}$

$t=2\frac{\ln 10}{\ln 2}\approx 6.6$ days

g) A radioactive substance decays so that at the end of 5 years only 70% of the substance remains. To the nearest tenth of a year, how long will it take until only 50% of the substance remains?

$.7\ast N=N\ast e^{5\ast k}$

$k=-7.133\,5\times 10^{-2}$

$.5\ast N=N\ast e^{\left( -7.1335 \times 10^{-2}\right) \ast t}$

$t\approx 9.7$ years

h) Assume that the cost of a car is $30,000. With continuous compounding in effect, find the number of years it would take to double the cost of the car at an annual inflation rate of 2.4%. (Round the answer to the nearest hundredth.)

$60000=30000\ast e^{.024\ast t}$

$t=28.88$ years

9. a) $108^{\circ }=\frac{\pi }{180}\ast 108= \frac{3}{5} \pi$ radians          b) $720^{\circ }=\frac{\pi }{180}\ast 720= 4\pi$ radians

c) $315^{\circ }=\frac{\pi }{180}\ast 315=\frac{7}{4}\pi$ radians        

d) $-135^{\circ }=\frac{\pi }{180}\ast \left( -135\right) = -\frac{3}{4}\pi$ radians

e) $77^{\circ }30^{\prime }=\frac{\pi }{180}\ast \left( 77.5\right) \approx 1.35$ radians                  f) $0^{\circ }=\frac{\pi }{180}\ast 0=0$ radians

g) $-335^{\circ }18^{\prime }35^{^{\prime \prime }}= -\frac{\pi }{180}\ast \left( 3... ...{18}{60}+\frac{35}{3600}\right) = -\frac{241\,423}{129\,600}\pi \approx -5.852$ radians

10. a) $-\frac{\pi }{3}$ radians $= -\frac{\pi }{3}\ast \frac{180}{\pi }= -60^{\circ }$

b) $\frac{\pi }{24}$ radians $=\frac{\pi }{24}\ast \frac{180}{\pi }= \frac{15}{2}=7.5^{\circ }$

c) $\frac{11\pi }{6}$ radians $=\frac{11\ast \pi }{6}\ast \frac{180}{\pi }=330^{\circ }$         d) $\frac{3\pi }{2}$ radians $=\frac{3\ast \pi }{2}\ast \frac{180}{\pi } =270^{\circ }$

e) $3$ radians $=3\ast \frac{180}{\pi }=\left( \frac{540}{\pi }\right) ^{\circ } =\approx 171.89^{\circ }$         f) $0$ radians $=0\ast \frac{180}{\pi }=0^{\circ }$

11. a) Find the radian measure of the central angle which subtends an arc of length 12 in a circle of radius 16.

$\theta =\frac{12}{16}=\frac{3}{4}= .75$ radians

b) Find the arc length subtended by a central angle of $\frac{\pi }{8}$ radians in a circle of radius 4.

$s=\frac{\pi }{8}\ast 4=\frac{\pi }{2}$

c) Find the radius of the circle in which an angle of 3 radians cuts off an arc of length 6.

$3=\frac{6}{r}$

$r=2$

d) Given a circle of radius 2 inches, determine the radian measure of the central angle which cuts off an arc of $\frac{\pi }{9}$ inches.

$\theta =\frac{\frac{\pi }{9}}{2}=\frac{\pi }{18}$

12. If $\theta$ is an angle in the first quadrant such that $\sin (\theta )=0.8=\frac{4}{5}$, find the exact values of each of the other fivetrigonometric functions of $\theta$.

a) $\cos (\theta )=\cos (\arcsin (.8))=.6=\frac{3}{5}$ Note: $\arcsin \left( \theta \right)$ is the same as $\sin ^{-1}\left( \theta \right)$

b) $\tan (\theta )=\tan \left( \arcsin (.8)\right) =\frac{4}{3}$ c) $\cot (\theta )=\frac{1}{\tan \left( \theta \right) }=\frac{3}{4}$

d) $\sec (\theta )=\frac{1}{\cos \left( \theta \right) }=\frac{5}{3}$                                          e) $\csc (\theta )=\frac{1}{\sin \left( \theta \right) }=\frac{5}{4}$

13. Given that the point $(-3,6)$ lies on the terminal side of an angle in standard position whose measure is $\theta$ radians, find the exact value of each of the following:

a) $\sin (\theta )=\frac{6}{\sqrt{9+36}}=\frac{6}{\sqrt{45}}= \frac{6}{3\sqrt{5}}=\frac{2}{\sqrt{5}}=\frac{2\sqrt{5}}{5}$

b) $\tan (\theta )=\frac{6}{-3}=-2$

14.

	$\theta$	$\sin \left( \theta \right)$	$\cos \left( \theta \right)$	$\tan \left( \theta \right)$	$\cot \left( \theta \right)$	$\sec \left( \theta \right)$	$\csc \left( \theta \right)$
a)	$\frac{3\pi }{4}$	$\frac{\sqrt{2}}{2}$	$-\frac{\sqrt{2}}{2}$	$-1$	$-1$	`$-\sqrt{2}$	$\sqrt{2}$
b)	$-\frac{\pi }{6}$	$-\frac{1}{2}$	$\frac{\sqrt{3}}{2}$	$-\frac{%% \sqrt{3}}{3}$	$-\sqrt{3}$	$\frac{2\sqrt{3}}{3}$	$-2$
c)	$240^{\circ }$	$-\frac{\sqrt{3}}{2}$	$-\frac{1}{2}$	$\sqrt{3}$	$\frac{\sqrt{3}}{3}$	$-2$	$-\frac{2\sqrt{3}}{3}$
d)	$-3\pi$	$0$	$-1$	$0$	n.d.	$-1$	n.d.
e)	$840^{\circ }$	$\frac{\sqrt{3}}{2}$	$-\frac{1}{2}$	$-\sqrt{3}$	$-\frac{\sqrt{3}}{3}$	$-2$	$\frac{2\sqrt{3}}{3}$
f)	$-\frac{\pi }{2}$	$-1$	$0$	n.d.	$0$	n.d.	$-1$
g)	$\frac{3\pi }{4}$	$\frac{\sqrt{2}}{2}$	$-\frac{\sqrt{2}}{2}$	$-1$	$-1$	$-\sqrt{2}$	$\sqrt{2}$
h)	$-210^{\circ }$	$\frac{1}{2}$	$-\frac{\sqrt{3}}{2}$	$-\frac{\sqrt{3}}{3}$	$-\sqrt{3}$	$-\frac{2\sqrt{3}}{3}$	$2$
i)	$\theta$	$\frac{12}{13}$	$-\frac{5}{13}$	$-\frac{12}{5}$	$-\frac{5}{12}$	$-\frac{13}{5}$	$\frac{13}{12}$

15. a) $\csc (\frac{\pi }{4})=\sqrt{2}$                          b) $\tan (\frac{\pi }{4})=1$

c) $\cos (240^{\circ })= -\frac{1}{2}$                          d) $\sec (180^{\circ })= -1$

e) The angle in the third quadrant whose secant is $-\sqrt{2}$ is $225^{\circ }$

16. a) $\left( \cos \left( -\frac{\pi }{2}\right) ,\sin \left( -%% \frac{\pi }{2}\right) \right) = \left( 0,-1\right)$

b) $\left( \cos \left( -\frac{2\pi }{3}\right) ,\sin \left( -\frac{2\pi }{3}\right) \right) =\left( -\frac{1}{2},-\frac{1}{2}\sqrt{3} \right)$

c) $\left( \cos \left( \frac{3\pi }{4}\right) ,\sin \left( \frac{3\pi }{4}%% \right) \right) =\left( -\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2} \right)$

17. a) $\cos (21^{\circ }10^{\prime })\approx .9325$                          b) $\ \csc (83^{\circ })\approx 1.007\,5$

c) $\cot \left( 310^{\circ }\right) \approx -.8391$                          d) $\sin (-200^{\circ })\approx .3420$

e) $\tan \left( \frac{\pi }{8}\right) \approx 4142$

18. Find an angle, $\theta$, in the indicated region, for which the following holds:

a) $\cos (\theta )=-\frac{\sqrt{3}}{2}\qquad$(second quadrant); $\theta=150^{\circ }$

b) $\tan \left( \theta \right) =1\qquad$(third quadrant); $\theta=225^{\circ }$

c) $\sin (\theta )=-\frac{\sqrt{2}}{2}$ (fourth quadrant); $\theta=-45^{\circ }$

d) $\cot \left( \theta \right) =0$ ( $\pi <\theta <2\pi$); $\theta = \frac{3\pi }{2}$

e) $\sec \left( \theta \right)$ is not defined ( $\pi <\theta <2\pi$); $\theta = \frac{3\pi }{2}$

19. a) $\sec \left( \theta \right)$, where the terminal side of $\theta$ is in quadrant III and $\sin (\theta )=~-~\frac{3}{11}$:         $\sec \left( \theta \right) =-\frac{11}{\sqrt{112}}=-\frac{11}{4\sqrt{7}}= -\frac{11}{28}\sqrt{7}$

b) $\csc \left( \theta \right)$, where the terminal side of $\theta$ is in quadrant III and $\sin (\theta )=~-~\frac{3}{11}$: $\csc \left( \theta \right) =-\frac{11}{3}$

c) $\tan (\theta )$, where the terminal side of $\theta$ is in quadrant IV and $\cos \left( \theta \right) =\frac{5}{9}$: $\tan (\theta )=-%% \frac{\sqrt{56}}{5}=-\frac{2\sqrt{14}}{5}$

d) $\sec (\theta )$, where the terminal side of $\theta$ is in quadrant IV and $\sin \left( \theta \right) =r$: $\sec (\theta )=\frac{1}{\sqrt{1-r^{2}}}$

e) $\tan (\theta )$, where the terminal side of $\theta$ is in quadrant IV and $\cos \left( \theta \right) =~1~-~r^{2}$: $\tan (\theta )=-\frac{\sqrt{2r^{2}-r^{4}}}{1-r^{2}}$

Note that part e) was supposed to read
$\cos \left( \theta \right) =\sqrt{1-r^{2}}$, which would have resulted in a simpler answer.

20. a) $\cos (32^{\circ }20^{\prime })\approx .8450$                          b) $\cot (241^{\circ })\approx .5543$

c) The angle in the second quadrant whose sine is .8774; $\theta \approx 118.7^{\circ }$

d) The nonnegative angle less than $360^{\circ }$ in the third quadrant whose sine is -0.8290; $\theta \approx 236.0$

e) The nonnegative angle less than $360^{\circ }$ in the fourth quadrant whose sine is -0.1569; $\theta \approx 351.0$

21. In this problem, $C$ is a right angle and

a) $a=5$ and $b=12$;

$c=\sqrt{5^{2}+12^{2}}=13$; $\tan \left( A\right) =\frac{5}{13}$, so $A=\arctan \left( \frac{5}{12}\right) \approx 22.6^{\circ }$;

$B=\arctan \left( \frac{12}{5}\right) \approx 67.4^{\circ }$

b) $a=24$ and $c=25$;

$b=\sqrt{25^{2}-24^{2}}=7$; $\sin \left( A\right) =\frac{24}{25}$, so $A=\arcsin \left( \frac{24}{25}\right) \approx 73.7^{\circ }$;

$B=\arccos \left( \frac{24}{25}\right) \approx 16.3^{\circ }$

c) $a=10$ and $\ A=\frac{\pi }{6}$;

$B=\frac{\pi }{2}-\frac{\pi }{6}= \frac{\pi }{3}$; $sin \left( \frac{\pi }{6}\right) =\frac{a}{c}$, so $c=\frac{a}{\sin \left( \frac{\pi }{6}\right)}=\frac{10}{\frac{1}{2}}=20$;

$b=\sqrt{20^{2}-10^{2}}\approx 17.3$

d) $a=12$ and $B=63^{\circ }$; $\ \ A=90^{\circ }-63^{\circ}= 27^{\circ }$;

$\cos \left( 63^{\circ }\right) =\frac{a}{c}=\frac{12}{c}$,

so $c=\frac{12}{\cos \left( 63^{\circ }\right) }\approx 26.4$; $\tan \left(63^{\circ }\right) =\frac{b}{a}=\frac{6}{12}$,

so $b=12\ast \tan \left(63^{\circ }\right) \approx 23.6$

e) $c=16$ and $A=\frac{\pi }{4}$;

$B=\frac{\pi }{2}-\frac{\pi }{4}= \frac{\pi }{4}$; $\sin \left( \frac{\pi }{4}\right) =\frac{a}{c}=\frac{a}{16}$, so $a=16\ast \sin \left( \frac{\pi }{4}\right) \approx 11.3$;

similarly, $b\approx 11.3$

f) $b=30$ and $A=30^{\circ }$; $B=90^{\circ }-30^{\circ}=60^{\circ }$;

$\tan \left( 30^{\circ }\right) =\frac{a}{b}=\frac{a}{30}$, so $a=30\ast \tan \left( 30^{\circ }\right) \approx 17.3$; $\sec \left( 30^{\circ }\right) =\frac{c}{b}=\frac{c}{30}$,

so $c=30\ast \sec \left(30^{\circ }\right) \approx 34.6$

22. a) $a=8$, $b=13$ and $c=15$;

$c^{2}=a^{2}+b^{2}-2\ast a\ast b\ast \cos \left( C\right)$; $15^{2}=8^{2}+13^{2}-2\ast 8\ast 13\ast \cos \left( C\right)$;

$\cos \left( C\right) =\frac{225-64-169}{-16\ast 13}= \frac{-8}{-16\ast 13}\approx 3.8462\times 10^{-2}$;

$C=\arccos \left( 3.8462\times 10^{-2}\right) \approx 87.8^{\circ }$

$b^{2}=a^{2}+c^{2}-2\ast a\ast c\ast \cos \left( B\right)$; $13^{2}=8^{2}+15^{2}-2\ast 8\ast 15\ast \cos \left( B\right)$;

$\cos \left( B\right) =\frac{169-64-225}{-16\ast 15}=\frac{-120}{-16\ast 15}=.5$; $C=\arccos \left( .5\right) =60^{\circ }$

$a^{2}=b^{2}+c^{2}-2\ast b\ast c\ast \cos \left( A\right)$; $8^{2}=13^{2}+15^{2}-2\ast 13\ast 15\ast \cos \left( A\right)$;

$\cos \left( A\right) =\frac{64-169-225}{-26\ast 15}=\frac{-330}{-26\ast 15}%% \approx .8462$; $A=\arccos \left( .8462\right) \approx 32.2^{\circ }$.

b) $a=5$, $b=9$ and $C=51^{\circ }$;

$c^{2}=a^{2}+b^{2}-2\ast a\ast b\ast \cos \left( C\right) =25+81-90\cos \left( 51^{\circ }\right) = 49.3611$;

$c\approx \sqrt{49.3611}\approx 7.0$;

$b^{2}=a^{2}+c^{2}-2\ast a\ast c\ast \cos \left( B\right)$; $81=25+49.3611-2\ast 5\ast 7.0\ast \cos \left( B\right)$;

$\cos \left( B\right) =\frac{81-25-49.3611}{-70}=-9.4841\times 10^{-2}$;

$B=\arccos \left( -9.4841\times 10^{-2}\right) \approx 95.4^{\circ }$;

$a^{2}=b^{2}+c^{2}-2\ast b\ast c\ast \cos \left( A\right)$; $25=81+49.3611-2\ast 9\ast 7\ast \cos \left( A\right)$;

$\cos \left( A\right) =\frac{25-81-49.3611}{-126}=.8362$; $A=\arccos \left( .8362\right) \approx 33.3^{\circ }$.

Note that the angles of the triangle do not seem to quite add up to $%% 180^{\circ }$. This is due to round-off error. If we use the Law of Sines to find $A$ and $B$, the round-off error is even greater.

c) $\ a=3$, $b=7$ and $C=105^{\circ }$

$c^{2}=a^{2}+b^{2}-2\ast a\ast b\ast \cos \left( C\right) =9+49-2\ast 3\ast 7\ast \cos \left( 105^{\circ }\right) \approx 68.8704$;

$c=\sqrt{68.8704}\approx 8.3$

$b^{2}=a^{2}+c^{2}-2\ast a\ast c\ast \cos \left( B\right)$; $49=9+68.8704-2\ast 3\ast 8.3\ast \cos \left( B\right)$;

$\cos \left( B\right) =\frac{49-9-68.8704}{-49.8}=.57973$; $B=\arccos \left( .57973\right) \approx 54.6^{\circ }$;

$a^{2}=b^{2}+c^{2}-2\ast b\ast c\ast \cos \left( A\right)$; $9=49+68.8704-2\ast 7\ast 8.3\ast \cos \left( A\right)$;

$\cos \left( A\right) =\frac{9-49-68.8704}{-14\ast 8.3}=.93692$; $A=\arccos \left( .93692\right) \approx 20.5^{\circ }$.

d) $a=12,$ $A=30^{\circ }$ and $B=72^{\circ }$;          $C=78^{\circ }$;

$\frac{a}{\sin \left( A\right) }=\frac{b}{\sin \left( B\right) }$ so $\frac{12}{\sin \left( 30^{\circ }\right) }= \frac{b}{\sin \left( 72^{\circ}\right) }$. Thus $b=\frac{12\sin \left( 72^{\circ }\right) }{\sin \left( 30^{\circ }\right) } \approx 22.8$;

$\frac{a}{\sin \left( A\right) }=\frac{c}{\sin \left( C\right) }$ so $\frac{12}{\sin \left( 30^{\circ }\right) }= \frac{c}{\sin \left( 78^{\circ}\right) }$. Thus $c=\frac{12\sin \left( 78^{\circ }\right) }{\sin \left( 30^{\circ }\right) }\approx 23.5$.

e) $a=15$, $B=45^{\circ }$ and $C=65^{\circ }$;        $A=70^{\circ }$;

$\frac{a}{\sin \left( A\right) }=\frac{b}{\sin \left( B\right) }$ so $\frac{15}{\sin \left( 70^{\circ }\right) }=\frac{b}{\sin \left( 45^{\circ }\right) }$. Thus $b=\frac{15\sin \left( 45^{\circ }\right) }{\sin \left( 70^{\circ }\right) }\approx 11.3$;

$\frac{a}{\sin \left( A\right) }=\frac{c}{\sin \left( C\right) }$ so $%% \frac{15}{\sin \left( 70^{\circ }\right) }=\frac{c}{\sin \left( 65^{\circ }\right) }$. Thus $c=\frac{15\sin \left( 65^{\circ }\right) }{\sin \left( 70^{\circ }\right) }\approx 14.5$.

f) $a=10$, $b=16$ and $A=60^{\circ }$;

$\frac{\sin \left( A\right) }{a}=\frac{\sin \left( B\right) }{b}$ so $\frac{\sin \left( 60^{\circ }\right) }{10}= \frac{\sin \left( B\right) }{16}$. Thus $\sin \left( B\right) =\frac{16\sin \left( 60^{\circ }\right) }{10} \approx 1.3856$;

Since the sine of an angle cannot exceed $1$, no such triangle is possible.

g) $a=16$, $b=11$ and $A=110^{\circ }$;

$\frac{\sin \left( A\right) }{a}=\frac{\sin \left( B\right) }{b}$ so $\frac{\sin \left( 110^{\circ }\right) }{16}=\frac{\sin \left( B\right) }{11}$. $\sin \left( B\right) =\frac{11\sin \left( 110^{\circ }\right) }{16}=.64604$

$B=\arcsin \left(.64604 \right) \approx 40.2^{\circ }$; therefore, $C\approx 180-110-40.2=29.8^{\circ }$; finally,

$c^{2}=a^{2}+b^{2}-2\ast a\ast b\ast \cos \left( C\right) =256+121-352\ast \cos \left( 29.8^{\circ }\right) =71.547$;

so $c\approx 8.5$.

h) $a=12.20$, $b=4.70$, and $C=120^{\circ }45$.

$c^{2}=a^{2}+b^{2}-2\ast a\ast b\ast \cos \left( C\right)$

$\qquad =\left( 12.2\right) ^{2}+\left( 4.7\right) ^{2}-2\ast 12.2\ast 4.7\ast \cos \left( 120.75^{\circ }\right) \approx 229.57$; so

$c\approx \sqrt{229.57}\approx 15.15$;

$\frac{\sin \left( A\right) }{a}=\frac{\sin \left( B\right) }{b}$ so $\frac{\sin \left( A\right) }{12.2}= \frac{\sin \left( 120.75^{\circ}\right) }{15.15}$; $\sin \left( A\right) =\frac{12.2\ast \sin \left(120.75^{\circ }\right) }{15.15} \approx .69206$;

$A=\arcsin \left( .69206\right) \approx 43.8^{\circ }$;

therefore, $B\approx 180-120.75-43.8\approx 15.45^{\circ }$.

23. a) A man looking through a telescope from the window of his 47th floor apartment sees a murder being committee on a lower floor in a building which is 1000 feet away. If the angle of declination of his telescope is set to $14^{\circ }20^{\prime }$, and if each story of both buildings is about 10 feet high, on what floor of the building did the police discover the body?

$\tan \left( 14.3333^{\circ }\right) =\frac{h}{1000}$; $h\approx 1000\ast \tan \left( 14.3333^{\circ }\right) \approx 255.52$ feet $\approx 25.5$ stories. Thus the murder took place on the 21$^{st}$ or maybe the 22$^{nd}$ floor.

b) A man on the ground following the flight of a baloon finds that the balloon is momentarily obscured from view by a street sign. If the sign is 24 feet above the ground and if the man is standing 32 feet from the base of the sign's support, what is the angle of elevation of the balloon at the moment that it disappears from view?

$\tan \left( \theta \right) =\frac{24}{32}=.75$; so $\theta =\arctan \left( .75\right) \approx 36.9^{\circ }$.

c) A boy runs 120 feet in a straight line, then veers to the left $20^{\circ }$ and runs 60 feet in this new direction. To the nearest foot, how far is he now from his starting point?

$c^{2}=a^{2}+b^{2}-2\ast a\ast b\ast \cos \left( C\right) =120^{2}+60^{2}-2\ast 120\ast 60\ast \cos \left( 160^{\circ }\right) =31532$, so $c=\sqrt{31532}\approx 178$ feet.

Note, when he veers left by $20^{\circ }$, the interior angle of the triangle is $160^{\circ }$.

d) Two towns Mudville and Crudville, on the same side of a straight river, are 20 miles apart. From Mudville, facing Crudville, Dudville, which is on the other bank of the river, forms an angle of $35^{\circ }$. From Crudville, facing Mudville, Dudville forms an angle of $47^{\circ }$. To the nearest tenth of a mile, how far, as the crow flies, is it from Mudville to Dudville?

By the Law of Sines, $\frac{d}{\sin \left( 47^{\circ }\right) } =\frac{20}{\sin \left( 35^{\circ }\right) }$, so $d=$ $\frac{20\ast \sin \left( 47^{\circ }\right) }{\sin \left( 35^{\circ }\right) }\approx 25.5$ miles.

Part ii of the original problem (which has been eliminated) was not a well-posed question. We can answer it, however, if we assume that both Mudville and Crudville also lie on the banks of the river.

First note that the angle at Mudville (i.e., Crudville-Mudville-Dudville) is $180-35-47=98^{\circ }$.

If we drop a perpendicular from Dudville to the opposite side of the river, then $\sin \left( 98^{\circ }\right) =\frac{r}{25.5}$, so $r\approx 25.3$ and the river is approximately 25.3 miles wide.

e) From the roof of one building, an observer can see the top of another building with an angle of elevation $29^{\circ }$. If the second building is 500 yards from the first, how much taller is the second building than the first?

$\tan \left( 29^{\circ }\right) =\frac{h}{500}$; so $h=500\ast \tan \left( 29^{\circ }\right) \approx 277.2$ yards. Thus the second building is approximately 277.2 yards taller than the first.