Chapter 10

The altitude extends from any vertex to the opposite side or to the line containing the opposite side at a 90° angle.

When the known values are the side opposite the missing angle and another side and its opposite angle.

A triangle with two given sides and a non-included angle.

$\beta =\mathrm{72°},a\approx 12.0,b\approx 19.9$

$\gamma =\mathrm{20°},b\approx 4.5,c\approx 1.6$

$b\approx 3.78$

$c\approx 13.70$

one triangle, $\alpha \approx \mathrm{50.3°},\beta \approx \mathrm{16.7°},a\approx 26.7$

two triangles, $\gamma \approx \mathrm{54.3°},\beta \approx \mathrm{90.7°},b\approx 20.9$ or ${\gamma }^{\prime }\approx \mathrm{125.7°},{\beta }^{\prime }\approx \mathrm{19.3°},b^{\prime }\approx 6.9$

two triangles, $\beta \approx \mathrm{75.7°},\gamma \approx \mathrm{61.3°},b\approx 9.9$ or ${\beta }^{\prime }\approx \mathrm{18.3°},{\gamma }^{\prime }\approx \mathrm{118.7°},b^{\prime }\approx 3.2$

two triangles, $\alpha \approx \mathrm{143.2°},\beta \approx \mathrm{26.8°},a\approx 17.3$ or ${\alpha }^{\prime }\approx \mathrm{16.8°},{\beta }^{\prime }\approx \mathrm{153.2°},a^{\prime }\approx 8.3$

no triangle possible

$A\approx \mathrm{47.8°}$ or $A^{\prime }\approx \mathrm{132.2°}$

$8.6$

$370.9$

$12.3$

$12.2$

$16.0$

$\mathrm{29.7°}$

$x=\mathrm{76.9°}\text{or}x=\mathrm{103.1°}$

$\mathrm{110.6°}$

$A\approx 39.4,C\approx 47.6,BC\approx 20.7$

$57.1$

$42.0$

$430.2$

$10.1$

$AD\approx 13.8$

$AB\approx 2.8$

$L\approx 49.7,N\approx 56.3,LN\approx 5.8$

51.4 feet

The distance from the satellite to station $A$ is approximately 1716 miles. The satellite is approximately 1706 miles above the ground.

2.6 ft

5.6 km

371 ft

5936 ft

24.1 ft

19,056 ft²

445,624 square miles

8.65 ft²

two sides and the angle opposite the missing side.

$s$ is the semi-perimeter, which is half the perimeter of the triangle.

The Law of Cosines must be used for any oblique (non-right) triangle.

11.3

34.7

26.7

257.4

not possible

95.5°

26.9°

$B\approx \mathrm{45.9°},C\approx \mathrm{99.1°},a\approx 6.4$

$A\approx \mathrm{20.6°},B\approx \mathrm{38.4°},c\approx 51.1$

$A\approx \mathrm{37.8°},B\approx 43.8,C\approx \mathrm{98.4°}$

177.56 in²

0.04 m²

0.91 yd²

3.0

29.1

0.5

70.7°

77.4°

25.0

9.3

43.52

1.41

0.14

18.3

48.98

7.62

85.1

24.0 km

99.9 ft

37.3 miles

2371 miles

292.4 miles

65.4 cm²

468 ft²

For polar coordinates, the point in the plane depends on the angle from the positive x-axis and distance from the origin, while in Cartesian coordinates, the point represents the horizontal and vertical distances from the origin. For each point in the coordinate plane, there is one representation, but for each point in the polar plane, there are infinite representations.

Determine $\theta$ for the point, then move $r$ units from the pole to plot the point. If $r$ is negative, move $r$ units from the pole in the opposite direction but along the same angle. The point is a distance of $r$ away from the origin at an angle of $\theta$ from the polar axis.

The point $\left(-3,\frac{\pi }{2}\right)$ has a positive angle but a negative radius and is plotted by moving to an angle of $\frac{\pi }{2}$ and then moving 3 units in the negative direction. This places the point 3 units down the negative y-axis. The point $\left(3,-\frac{\pi }{2}\right)$ has a negative angle and a positive radius and is plotted by first moving to an angle of $-\frac{\pi }{2}$ and then moving 3 units down, which is the positive direction for a negative angle. The point is also 3 units down the negative y-axis.

$\left(-5,0\right)$

$\left(-\frac{3\sqrt{3}}{2},-\frac{3}{2}\right)$

$\left(2\sqrt{5},0.464\right)$

$\left(\sqrt{34},5.253\right)$

$\left(8\sqrt{2},\frac{\pi }{4}\right)$

$r=4\csc \theta$

$r=\sqrt[3]{\frac{sin\theta }{2co{s}^4\theta }}$

$r=3\cos \theta$

$r=\frac{3\sin \theta }{\cos \left(2\theta \right)}$

$r=\frac{9\sin \theta }{{\cos }^2\theta }$

$r=\sqrt{\frac{1}{9\cos \theta \sin \theta }}$

$x^2+y^2=4x$ or $\frac{{\left(x-2\right)}^2}{4}+\frac{y^2}{4}=1;$ circle

$3y+x=6;$ line

$y=3;$ line

$xy=4;$ hyperbola

$x^2+y^2=4;$ circle

$x-5y=3;$ line

$\left(3,\frac{3\pi }{4}\right)$

$\left(5,\pi \right)$

$r=\frac{6}{5\cos \theta -\sin \theta }$

$r=2\sin \theta$

$r=\frac{2}{\cos \theta }$

$r=3\cos \theta$

$x^2+y^2=16$

$y=x$

$x^2+{\left(y+5\right)}^2=25$

$\left(1.618,-1.176\right)$

$\left(10.630,\mathrm{131.186°}\right)$

$\left(2,3.14\right)or\left(2,\pi \right)$

A vertical line with $a$ units left of the y-axis. 

A horizontal line with $a$ units below the x-axis.

Symmetry with respect to the polar axis is similar to symmetry about the $x$-axis, symmetry with respect to the pole is similar to symmetry about the origin, and symmetric with respect to the line $\theta =\frac{\pi }{2}$ is similar to symmetry about the $y$-axis.

Test for symmetry; find zeros, intercepts, and maxima; make a table of values. Decide the general type of graph, cardioid, limaçon, lemniscate, etc., then plot points at $\theta =0,\frac{\pi }{2},$ $\pi$ and  $\frac{3\pi }{2},$ and sketch the graph.

The shape of the polar graph is determined by whether or not it includes a sine, a cosine, and constants in the equation.

symmetric with respect to the polar axis

symmetric with respect to the polar axis, symmetric with respect to the line $\theta =\frac{\pi }{2},$ symmetric with respect to the pole

no symmetry

no symmetry

symmetric with respect to the pole

circle

cardioid

cardioid

one-loop/dimpled limaçon

one-loop/dimpled limaçon

inner loop/two-loop limaçon

inner loop/two-loop limaçon

inner loop/two-loop limaçon

lemniscate

lemniscate

rose curve

rose curve

Archimedes’ spiral

Archimedes’ spiral

They are both spirals, but not quite the same.

Both graphs are curves with 2 loops. The equation with a coefficient of $\theta$ has two loops on the left, the equation with a coefficient of 2 has two loops side by side. Graph these from 0 to $4\pi$ to get a better picture.

When the width of the domain is increased, more petals of the flower are visible.

The graphs are three-petal, rose curves. The larger the coefficient, the greater the curve’s distance from the pole.

The graphs are spirals. The smaller the coefficient, the tighter the spiral.

$\left(4,\frac{\pi }{3}\right),\left(4,\frac{5\pi }{3}\right)$

$\left(\frac{3}{2},\frac{\pi }{3}\right),\left(\frac{3}{2},\frac{5\pi }{3}\right)$

$\left(0,\frac{\pi }{2}\right),\left(0,\pi \right),\left(0,\frac{3\pi }{2}\right),\left(0,2\pi \right)$

$\left(\frac{\sqrt[4]{8}}{2},\frac{\pi }{4}\right),\left(\frac{\sqrt[4]{8}}{2},\frac{5\pi }{4}\right)$ and at $\theta =\frac{3\pi }{4},$ $\frac{7\pi }{4}$ since $r$ is squared

a is the real part, b is the imaginary part, and $i=\sqrt{-1}$

Polar form converts the real and imaginary part of the complex number in polar form using $x=r\cos \theta$ and $y=r\sin \theta .$

$z^n=r^n\left(\cos \left(n\theta \right)+i\sin \left(n\theta \right)\right)$ It is used to simplify polar form when a number has been raised to a power.

$5\sqrt{2}$

$\sqrt{38}$

$\sqrt{14.45}$

$4\sqrt{5}\mathrm{cis}\left(\mathrm{333.4°}\right)$

$2\mathrm{cis}\left(\frac{\pi }{6}\right)$

$\frac{7\sqrt{3}}{2}+i\frac{7}{2}$

$-2\sqrt{3}-2i$

$-1.5-i\frac{3\sqrt{3}}{2}$

$4\sqrt{3}\mathrm{cis}\left(\mathrm{198°}\right)$

$\frac{3}{4}\mathrm{cis}\left(\mathrm{180°}\right)$

$5\sqrt{3}\mathrm{cis}\left(\frac{17\pi }{24}\right)$

$7\mathrm{cis}\left(\mathrm{70°}\right)$

$5\mathrm{cis}\left(\mathrm{80°}\right)$

$5\mathrm{cis}\left(\frac{\pi }{3}\right)$

$125\mathrm{cis}\left(\mathrm{135°}\right)$

$9\mathrm{cis}\left(\mathrm{240°}\right)$

$\mathrm{cis}\left(\frac{3\pi }{4}\right)$

$3\mathrm{cis}\left(\mathrm{80°}\right),3\mathrm{cis}\left(\mathrm{200°}\right),3\mathrm{cis}\left(\mathrm{320°}\right)$

$2\sqrt[3]{4}\mathrm{cis}\left(\frac{2\pi }{9}\right),2\sqrt[3]{4}\mathrm{cis}\left(\frac{8\pi }{9}\right),2\sqrt[3]{4}\mathrm{cis}\left(\frac{14\pi }{9}\right)$

$2\sqrt{2}\mathrm{cis}\left(\frac{7\pi }{8}\right),2\sqrt{2}\mathrm{cis}\left(\frac{15\pi }{8}\right)$

$3.61e^{-0.59i}$

$-2+3.46i$

$-4.33-2.50i$

A pair of functions that is dependent on an external factor. The two functions are written in terms of the same parameter. For example, $x=f\left(t\right)$ and $y=f\left(t\right).$

Choose one equation to solve for $t,$ substitute into the other equation and simplify.

Some equations cannot be written as functions, like a circle. However, when written as two parametric equations, separately the equations are functions.

$y=-2+2x$

$y=3\sqrt{\frac{x-1}{2}}$

$x=2e^{\frac{1-y}{5}}$ or $y=1-5ln\left(\frac{x}{2}\right)$

$x=4\log \left(\frac{y-3}{2}\right)$

$x={\left(\frac{y}{2}\right)}^3-\frac{y}{2}$

$y=x^3$

${\left(\frac{x}{4}\right)}^2+{\left(\frac{y}{5}\right)}^2=1$

$y^2=1-\frac{1}{2}x$

$y=x^2+2x+1$

$y={\left(\frac{x+1}{2}\right)}^3-2$

$y=-3x+14$

$y=x+3$

$\{\begin{array}{l}x\left(t\right)=t\hfill \\ y\left(t\right)=2\sin t+1\hfill \end{array}$

$\{\begin{array}{l}x\left(t\right)=\sqrt{t}+2t\hfill \\ y\left(t\right)=t\hfill \end{array}$

$\{\begin{array}{l}x\left(t\right)=4\cos t\hfill \\ y\left(t\right)=6\sin t\hfill \end{array};$ Ellipse

$\{\begin{array}{l}x\left(t\right)=\sqrt{10}\cos t\hfill \\ y\left(t\right)=\sqrt{10}\sin t\hfill \end{array};$ Circle

$\{\begin{array}{l}x\left(t\right)=-1+4t\hfill \\ y\left(t\right)=-2t\hfill \end{array}$

$\{\begin{array}{l}x\left(t\right)=4+2t\hfill \\ y\left(t\right)=1-3t\hfill \end{array}$

yes, at $t=2$

answers may vary: $\{\begin{array}{l}x\left(t\right)=t-1\hfill \\ y\left(t\right)=t^2\hfill \end{array}\text{ and }\{\begin{array}{l}x\left(t\right)=t+1\hfill \\ y\left(t\right)={\left(t+2\right)}^2\hfill \end{array}$

answers may vary: , $\{\begin{array}{l}x\left(t\right)=t\hfill \\ y\left(t\right)=t^2-4t+4\hfill \end{array}\text{ and }\{\begin{array}{l}x\left(t\right)=t+2\hfill \\ y\left(t\right)=t^2\hfill \end{array}$

plotting points with the orientation arrow and a graphing calculator

The arrows show the orientation, the direction of motion according to increasing values of $t.$

The parametric equations show the different vertical and horizontal motions over time.

There will be 100 back-and-forth motions.

Take the opposite of the $x\left(t\right)$ equation.

The parabola opens up.

$\{\begin{array}{l}x\left(t\right)=5\cos t\\ y\left(t\right)=5\sin t\end{array}$

$a=4,$ $b=3,$ $c=6,$ $d=1$

$a=4,$ $b=2,$ $c=3,$ $d=3$

The $y$-intercept changes.

$y\left(x\right)=-16{\left(\frac{x}{15}\right)}^2+20\left(\frac{x}{15}\right)$

$\{\begin{array}{l}x(t)=64t\cos \left(52°\right)\\ y(t)=-16t^2+64t\sin \left(52°\right)\end{array}$