# Math Help - Does anyone know how to get started on these questions?

1. ## Does anyone know how to get started on these questions?

Hi there trigonometry is not my subject and I was wondering if anyone knew how to get started on these questions. I've made some attempts but I don't think they were very successful. If anyone can help I would greatly appreciate it. Below are the questions.

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 degrees from the direction of travel. After the ship continues in the same direction for 575 m, the stature appears to be at an angle of 68 degrees from the direction of travel. How far from the statue is the ship at the later time.

On a 520 m hole, a golfer’s tee shot goes 175m and 17 degrees to the right of the direct path to the flag. The second shot flies 15 degrees to the right of the new direction toward the flag, but only travels 150 m. How far is the golf ball from the flag?

Two trappers travel 8.2 km on a bearing of 40 degrees and then, 6.3 km on a bearing of 160 degrees. What distance should they travel to return directly to their starting point.

Thanks,

2. 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: . $\angle CAB = 20^o$

The ship sails to B: . $AB \:=\:575\text{ m}$
. . and sights the statue again: . $\angle CBD = 68^o$

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

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

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

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

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

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