Hello, katelynn!

I'll get you started with the first one . . .

As a cruise ship arrives in a new port of call, passengers are invited to admire

a particular famous statue at an angle of 20° from the direction of travel.

After the ship continues in the same direction for 575 m, the statue

appears to be at an angle of 68° from the direction of travel.

How far from the statue is the ship at the later time?

Of course, the first step is to make a sketch . . . Code:

C
*
* *
*48°*
* *
* * a
* *
* *
* 20° 112° * 68°
* - - - - - - - * - - - - D
A 575 B

The ship is at A and sights the statue at C: .$\displaystyle \angle CAB = 20^o$

The ship sails to B: .$\displaystyle AB \:=\:575\text{ m}$

. . and sights the statue again: .$\displaystyle \angle CBD = 68^o$

We want the distance: $\displaystyle a \:=\:BC$

Since $\displaystyle \angle CBD = 68^o$, then $\displaystyle \angle CBA = 112^o$

In $\displaystyle \Delta ABC:\;\angle C + 20^o + 112^o \:=\:180^o\quad\Rightarrow\quad \angle C \:=\: 48^o$

Law of Sines: .$\displaystyle \frac{a}{\sin A} \:=\:\frac{c}{\sin C}$

We have: .$\displaystyle \frac{a}{\sin 20^o} \:=\:\frac{575}{\sin48^o}$

. . Therefore: .$\displaystyle a \:=\:\frac{575\sin20^o}{\sin48^o} \;\approx\;264.6\text{ m}$