$r=sqrt{x^2+y^2}$

$cos theta=dfrac{x}{r}$

$sin theta=dfrac{y}{r}$

$tan theta=dfrac{y}{x}$

$sectheta=dfrac{1}{costheta}=dfrac{r}{x}$

$csctheta=dfrac{1}{sin theta}=dfrac{r}{y}$

$$
cottheta=dfrac{1}{tantheta}=dfrac{x}{y}
$$

$r=sqrt{(-3)^2+0^2}=3$

i) $costheta=dfrac{-3}{3}=-1$

$sin theta=dfrac{0}{3}=0$

$tantheta=dfrac{0}{-3}=0$

$sectheta=dfrac{1}{costheta}=-1$

$csctheta=dfrac{1}{0}=text{undefined}$

$cottheta=dfrac{1}{0}=text{undefined}$

ii) The principal angle is $180^circ$ and related acute angle is $0^circ$.

$r=sqrt{(-8)^2+(-6)^2}=sqrt{100}=10$

$cos theta=dfrac{-8}{10}=-dfrac{4}{5}$

$sin theta=dfrac{-6}{10}=-dfrac{3}{5}$

$tan theta=dfrac{-6}{-8}=dfrac{3}{4}$

$sectheta=dfrac{1}{costheta}=-dfrac{5}{4}$

$csctheta=dfrac{1}{sintheta}=-dfrac{5}{3}$

$cottheta=dfrac{1}{tantheta}=dfrac{4}{3}$

The reference acute angle is

$beta=tan^{-1}left|dfrac{-6}{-8}right|=37^circ$

Since $theta$ is in quadrant 3, the principal angle is

$$
theta=180^circ+beta=180+37=217^circ
$$

$sectheta=-1$ , $csctheta$ and $cottheta$ are undefined

$theta=180^circ$, $beta=0^circ$, see sketch inside

b) $sin theta=-dfrac{3}{5}$, $costheta=-dfrac{4}{5}$ , $tantheta=dfrac{3}{4}$

$csctheta=-dfrac{5}{3}$ , $sectheta=-dfrac{5}{4}$ , $cottheta=dfrac{4}{3}$

$theta=217^circ$ , $beta=37^circ$, see sketch inside

1) Find the reference acute angle $beta$ using inverse trigonometric function as

$f(theta)=cimplies beta=f^{-1}(|c|)$

2) Based on the sign of $c$, determine the possible quadrants for $theta$ within the given range.

3) Find the principal angle $theta$ on those quadrants that has reference acute angle of $beta$

$sin theta0$ in QI and QIV

QI: $theta=beta=30^circ$

QIV: $theta=360^circ-beta=360^circ-30^circ=330^circ$

$cottheta<0$ in QII and QIV

QII: $theta=180^circ-beta=180^circ-45^circ=135^circ$

QIV: $theta=360^circ-beta=360^circ-45^circ=315^circ$

$sectheta=-2implies costheta=-dfrac{1}{2}implies beta=cos^{-1}left|-dfrac{1}{2}right|=60^circ$

$costheta<0$ in QII and QIII

QII: $theta=180^circ-beta=180^circ-60^circ=120^circ$

QII: $theta=180^circ+beta=180^circ+60^circ=240^circ$

b) $30^circ$ , $330^circ$

c) $135^circ$ , $315^circ$

d) $120^circ$ , $240^circ$

$sintheta=pmsqrt{1-cos^2theta}=pmsqrt{1-left(-dfrac{5}{13}right)}=pmsqrt{dfrac{144}{169}}=pmdfrac{12}{13}$

In QII, $sintheta>0$, thus, $sin theta=dfrac{12}{13}$

$$
sin thetacostheta=left(dfrac{12}{13}right)cdot left(-dfrac{5}{13}right)=-dfrac{60}{169}
$$

This is an identity so it would be true for all real $theta$

b) $1$

$=dfrac{sin^2 phi}{cos^2phi}+1$

$color{#c34632}text{Use} sin^2theta+cos^2theta=1$

$=dfrac{1-cos^2phi}{cos^2phi}+1$

$=dfrac{1}{cos^2phi}-1+1$

$=dfrac{1}{cos^2phi}$

$$
=sec^2phi
$$

$=1+dfrac{cos^2alpha}{sin^2alpha}$

$color{#c34632}text{Use} sin^2theta+cos^2theta=1$

$=1+dfrac{1-sin^2alpha}{sin^2alpha}$

$=1+dfrac{1}{sin^2alpha}-1$

$=dfrac{1}{sin^2alpha}$

$$
=csc^2alpha
$$

$$
x^2+y^2=r^2implies left(dfrac{x}{r}right)^2+left(dfrac{y}{r}right)^2=1implies sin^2theta+cos^2theta=1
$$

b) both are derived from $sin^2theta+cos^2theta=1$

$a^2=b^2+c^2-2bccos A$

$b^2=a^2+c^2-2accos B$

c) $dfrac{sin A}{a}=dfrac{sin B}{b}=dfrac{sin C}{c}$ or $dfrac{a}{sin A}=dfrac{b}{sin B}=dfrac{c}{sin C}$

b) $c^2=a^2+b^2-2abcos C$

$a^2=b^2+c^2-2bccos A$

$b^2=a^2+c^2-2accos B$

c) $dfrac{sin A}{a}=dfrac{sin B}{b}=dfrac{sin C}{c}$ or $dfrac{a}{sin A}=dfrac{b}{sin B}=dfrac{c}{sin C}$

Remember that $tan theta=dfrac{text{opposite side}}{text{adjacent side}}$

To find $w$, we can use tangent ratio to find $BD$ and $DC$, then add them together.

$tan 49^circ=dfrac{60.0}{BD}implies BD=dfrac{60.0}{tan 49^circ}=52.16$ m

$tan 53^circ=dfrac{60.0}{DC}implies DC=dfrac{60.0}{tan 53^circ}=45.21$ m

$w=BD+DC=52.16+45.21=97.4$ m

$tan 40^circ=dfrac{5.5}{GJ}implies GJ=dfrac{5.5}{tan 40^circ}=6.55$

$tan 48^circ=dfrac{5.5}{HJ}implies HJ=dfrac{5.5}{tan 48^circ}=4.95$

$w=GJ-HJ=6.55-4.95=1.6$ m

b) $1.6$ m

$bsin A < a < bimplies$ 2 triangles

$bsin A<b aimplies$ no triangle

$csin A<c aimplies$ no triangle