Question

name:____ period:__ date:____
6.
lk =
jk =
7.
wv =
uv =
8.
ed =
df =
9.
x =
y =
10.
a = 10
b = 5

Explanation:

Problem 6:

Step1: Solve for $LK$ using sine

$\sin(30^\circ)=\frac{9}{LK} \implies LK=\frac{9}{\sin(30^\circ)}=\frac{9}{\frac{1}{2}}=18$

Step2: Solve for $JK$ using cosine

$\cos(30^\circ)=\frac{JK}{LK} \implies JK=18\times\frac{\sqrt{3}}{2}=9\sqrt{3}$

Problem 7:

Step1: Solve for $WV$ using sine

$\sin(30^\circ)=\frac{WV}{12} \implies WV=12\times\frac{1}{2}=6$

Step2: Solve for $UV$ using cosine

$\cos(30^\circ)=\frac{UV}{12} \implies UV=12\times\frac{\sqrt{3}}{2}=6\sqrt{3}$

Problem 8:

Step1: Solve for $ED$ using sine

$\sin(30^\circ)=\frac{ED}{10\sqrt{3}} \implies ED=10\sqrt{3}\times\frac{1}{2}=5\sqrt{3}$

Step2: Solve for $DF$ using cosine

$\cos(30^\circ)=\frac{DF}{10\sqrt{3}} \implies DF=10\sqrt{3}\times\frac{\sqrt{3}}{2}=15$

Problem 9:

Step1: Solve for $x$ using cosine

$\cos(60^\circ)=\frac{x}{2\sqrt{2}} \implies x=2\sqrt{2}\times\frac{1}{2}=\sqrt{2}$

Step2: Solve for $y$ using sine

$\sin(60^\circ)=\frac{y}{2\sqrt{2}} \implies y=2\sqrt{2}\times\frac{\sqrt{3}}{2}=\sqrt{6}$

Problem 10:

Step1: Solve for $a$ using cosine

$\cos(60^\circ)=\frac{5\sqrt{3}}{a} \implies a=\frac{5\sqrt{3}}{\frac{1}{2}}=10\sqrt{3}$

Step2: Solve for $b$ using tangent

$\tan(60^\circ)=\frac{5\sqrt{3}}{b} \implies b=\frac{5\sqrt{3}}{\sqrt{3}}=5$

Answer:

1. $LK=18$, $JK=9\sqrt{3}$
2. $WV=6$, $UV=6\sqrt{3}$
3. $ED=5\sqrt{3}$, $DF=15$
4. $x=\sqrt{2}$, $y=\sqrt{6}$
5. $a=10\sqrt{3}$, $b=5$