Trigonometry : Height and Distance Question From a ship sailing due South-East at the rate of 5 miles an hour, light-house is observed to be $30^0$ North of East, and after 4 hours, it is seen due North ; find the distance of the light-house from the final position of the ship
My Solution

Let O be the position of ship  and L be the lighthouse. Given
∠LOB=$30^0$
When Ship reaches at point A, Lighthouse is observed at North.
t=4 hours
v=5 miles per hour
$$v=OA/t$$
$$OA=5×4=20 miles$$
In ΔOAB
$$sin60^0=AB/OA$$
$$√3/2=AB/20$$
$$AB=√3/2×20$$
$$=10√3  miles$$
Using Pythagoras theorem
$$OB^2=OA^2-AB^2$$
$$OB^2=20^2-(10√3)^2$$
$$OB^2=400-300$$
$$OB=10 miles$$
In ΔOBL
$$tan30^0=LB/OB$$
$$LB=10×1/√3=5.77 miles$$
Total Distance = LB + AB = 10√3+5.773=23.01 miles
But answer is 22.3 miles.
What is wrong in my solution ?
 A: $\angle AOB$ should be $45º$ instead of $60º$ as it is travelling due south-east.
Now since $\Delta OBA$ is isosceles, you should get that $AL = \frac{20}{\sqrt2} + \frac{20}{\sqrt2} \tan 30º = 22.3$ miles.
A: As mentioned by Toby Mak, $\angle AOB$ should be exactly $45$° since the ship is travelling south-east.
The length $OA$ calculated in your solution is correct, at $20$ miles.
Therefore, the following is true:

 $$\text {cos} (45\text {°}) = \frac {OB} {OA}$$

Since $OA$ is $20$ miles, $OB$ can be found as follows:

 $$OB = \text {cos} (45\text {°}) \cdot 20 \text { miles}$$
$$OB = 14.1 \text { miles}$$

$AB$ can be found in a similar manner:

 $$\text {sin} (45\text {°}) = \frac {AB} {OA}$$
$$AB = \text {sin} (45\text {°}) \cdot 20 \text { miles}$$
$$AB = 14.1 \text { miles}$$

Finally, $LB$ can be found using $OB$, as follows:

 $$\text {tan} (30\text {°}) = \frac {LB} {OB}$$
$$LB = \text {tan} (30\text {°}) \cdot 14.1 \text { miles}$$
$$LB = 8.2 \text { miles}$$

Thus, since $AL = AB + LB$, $AL$ can be found as follows:

 $$AL = 14.1 \text { miles} + 8.2 \text { miles} = 22.3 \text { miles}$$

I hope that helps!
