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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 enter image description here

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 ?

Sagar Rawal
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2 Answers2

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$\angle AOB$ 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.

Toby Mak
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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!

anipalur
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