# A boat is sailing due east parallel to the shoreline at a speed of 11 miles per hour. At a given time, the bearing to the lighthouse is S 70° E, and 30 minutes later the bearing is S 63° E (see figure). The lighthouse is located at the shoreline. help?

Jul 24, 2018

#### Answer:

$\text{Distance from shoreline} = 7.0 \left(1 \mathrm{dp}\right)$miles

#### Explanation:

There is no figure shown, but when taking bearings for navigation the direction to a point is stated as the number of degrees east or west of North or South.

The likely question that follows from the information given is "how far from the shoreline is the boat?"

We know that the boat is traveling parallel to the shoreline at $11$mph for $0.5$ hours, so it moves $5.5$ miles.

The 1st bearing to the lighthouse is more East than the 2nd bearing, so the boat is moving east, towards the lighthouse.

Let $a$ be the perpendicular distance to shore and $b$ be the distance along the shoreline to the lighthouse from the 2nd bearing position.

A not-to-scale drawing : $\tan \left(90 - 70\right) = \frac{a}{b + 5.5}$

#a=tan(20)*(b+5.5) ....(1)

and

$\tan \left(90 - 63\right) = \frac{a}{b}$

$b = \frac{a}{\tan} \left(27\right)$

Substitute for b in eqn (1)

$a = \tan \left(20\right) \cdot \left(\frac{a}{\tan} \left(27\right) + 5.5\right)$

$a = \left(a \cdot \tan \frac{20}{\tan} \left(27\right)\right) + \tan \left(20\right) \cdot 5.5$

$a - \left(a \cdot \tan \frac{20}{\tan} \left(27\right)\right) = \tan \left(20\right) \cdot 5.5$

$a \left(1 - \tan \frac{20}{\tan} \left(27\right)\right) = \tan \left(20\right) \cdot 5.5$

$a = \frac{\tan \left(20\right) \cdot 5.5}{1 - \tan \frac{20}{\tan} \left(27\right)}$

Evaluate (remember to use degrees):

$a = 7.0 \left(1 \mathrm{dp}\right)$miles

I hope this is what you needed.
It is easy to use the sine function to find the distances from the lighthouse from this if needed.