Question

find both variables
1.
2.
3.
4.
5.
6.
7.

1. jen is on the platform of her boat. she sights the top of a lighthouse at an angle of 30° as shown below. she knows the lighthouse is 50 meters high. how far is jen from the base of the lighthouse, in meters?
2. jeffrey wants to build a ramp to make it easier to load his lawn mower into the back of his truck. he drew the diagram below to help him design the ramp. what is the length of the ramp in feet?

jeffrey’s ramp design

Explanation:

Step1: Solve for a (Q1)

Use $\sin30^\circ=\frac{\text{opposite}}{\text{hypotenuse}}$
$\sin30^\circ=\frac{3}{a} \implies a=\frac{3}{\sin30^\circ}=\frac{3}{\frac{1}{2}}=6$

Step2: Solve for b (Q1)

Use $\tan30^\circ=\frac{\text{opposite}}{\text{adjacent}}$
$\tan30^\circ=\frac{3}{b} \implies b=\frac{3}{\tan30^\circ}=\frac{3}{\frac{1}{\sqrt{3}}}=3\sqrt{3}$

Step3: Solve for d (Q2)

Use $\sin30^\circ=\frac{d}{4}$
$d=4\sin30^\circ=4\times\frac{1}{2}=2$

Step4: Solve for c (Q2)

Use $\cos30^\circ=\frac{c}{4}$
$c=4\cos30^\circ=4\times\frac{\sqrt{3}}{2}=2\sqrt{3}$

Step5: Solve for f (Q3)

Use $\tan60^\circ=\frac{f}{8}$
$f=8\tan60^\circ=8\sqrt{3}$

Step6: Solve for e (Q3)

Use $\cos60^\circ=\frac{8}{e}$
$e=\frac{8}{\cos60^\circ}=\frac{8}{\frac{1}{2}}=16$

Step7: Solve for i (Q4)

Use $\tan60^\circ=\frac{10}{i}$
$i=\frac{10}{\tan60^\circ}=\frac{10}{\sqrt{3}}=\frac{10\sqrt{3}}{3}$

Step8: Solve for h (Q4)

Use $\sin60^\circ=\frac{10}{h}$
$h=\frac{10}{\sin60^\circ}=\frac{10}{\frac{\sqrt{3}}{2}}=\frac{20\sqrt{3}}{3}$

Step9: Solve for j (Q5)

Use $\tan60^\circ=\frac{j}{4}$
$j=4\tan60^\circ=4\sqrt{3}$

Step10: Solve for k (Q5)

Use $\cos60^\circ=\frac{4}{k}$
$k=\frac{4}{\cos60^\circ}=\frac{4}{\frac{1}{2}}=8$

Step11: Solve for t (Q6)

Use $\tan30^\circ=\frac{t}{9}$
$t=9\tan30^\circ=9\times\frac{1}{\sqrt{3}}=3\sqrt{3}$

Step12: Solve for r (Q6)

Use $\cos30^\circ=\frac{9}{r}$
$r=\frac{9}{\cos30^\circ}=\frac{9}{\frac{\sqrt{3}}{2}}=6\sqrt{3}$

Step13: Solve for p (Q7)

Use $\tan60^\circ=\frac{p}{5}$
$p=5\tan60^\circ=5\sqrt{3}$

Step14: Solve for v (Q7)

Use $\sin30^\circ=\frac{5}{v}$
$v=\frac{5}{\sin30^\circ}=\frac{5}{\frac{1}{2}}=10$

Step15: Solve Q8 distance

Use $\tan30^\circ=\frac{50}{x}$
$x=\frac{50}{\tan30^\circ}=\frac{50}{\frac{1}{\sqrt{3}}}=50\sqrt{3}$

Step16: Solve Q9 ramp length

Use $\sin30^\circ=\frac{2.5}{t}$
$t=\frac{2.5}{\sin30^\circ}=\frac{2.5}{\frac{1}{2}}=5$

Answer:

1. $a=6$, $b=3\sqrt{3}$
2. $d=2$, $c=2\sqrt{3}$
3. $f=8\sqrt{3}$, $e=16$
4. $i=\frac{10\sqrt{3}}{3}$, $h=\frac{20\sqrt{3}}{3}$
5. $j=4\sqrt{3}$, $k=8$
6. $t=3\sqrt{3}$, $r=6\sqrt{3}$
7. $p=5\sqrt{3}$, $v=10$
8. $50\sqrt{3}$ meters
9. 5 feet