From the top of a 325-ft lighthouse, the angle of depression to a ship in the ocean is 30°. How far is the ship from the base of the lighthouse? Round your answer to the nearest tenth of a foot.
+33616 Picture the triangle made by a line from the centre of the earth through to the top of the lighthouse (which I'll call c); a line from the top of the lighthouse to the ship (which I'll call b); and a line from the ship back to the centre of the earth (which I'll call a). I would prefer to show a diagram here, but I can't as we can't yet upload files!
Since the angle of depression is 30°, the angle, A, between lines b and c is 90° - 30° or 60°. We can write the cosine rule for this triangle as a2 = b2 + c2 -2bccos(A).
Now the average radius of the Earth is 6371km (according to Wikipedia), so line a is 6371km long. Line c is the radius of the Earth plus 325ft (=99.06m) so c is 6,371,000 + 99.06 = 6,371,099.06m. The only unknown in the cosine rule above is b, and we can solve for b to find two mathematical solutions. These are 6371km and 198.12m.
Clearly it is the smaller value we want here! So b = 198.12m.
Now we need angle B, the angle between lines a and c. We get this from the sine rule: sin(B)/b = sin(A)/a. So:
sin(B)=sin(60°)*198.12/6371000 = 2.693*10-5
Angle B = sin-1(2.693*10-5) or B = 2.693*10-5 radians (when an angle is very small its value in radians is approximately equal to its sine). The distance from the ship to the lighthouse along the curve of the Earth is just B*a where B is in radians. So $${\mathtt{distance}} = {\mathtt{2.693}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{\left(-{\mathtt{5}}\right)}{\mathtt{\,\times\,}}{\mathtt{6\,371\,000}} = {\mathtt{distance}} = {\mathtt{171.571\: \!03}}$$ metres
This is 562.9ft to the nearest tenth of a foot.
Alternatively
Since the distances are relatively small we could assume the surface of the Earth to be flat near the lighthouse, in which case the distance of the ship from the lighthouse is given by 325*tan(60°)ft or
$${\mathtt{distance}} = {\mathtt{325}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}{\left({\mathtt{60}}^\circ\right)} = {\mathtt{distance}} = {\mathtt{562.916\: \!512\: \!459\: \!925}}$$ feet
or 562.9ft.
+33616 Picture the triangle made by a line from the centre of the earth through to the top of the lighthouse (which I'll call c); a line from the top of the lighthouse to the ship (which I'll call b); and a line from the ship back to the centre of the earth (which I'll call a). I would prefer to show a diagram here, but I can't as we can't yet upload files!
Since the angle of depression is 30°, the angle, A, between lines b and c is 90° - 30° or 60°. We can write the cosine rule for this triangle as a2 = b2 + c2 -2bccos(A).
Now the average radius of the Earth is 6371km (according to Wikipedia), so line a is 6371km long. Line c is the radius of the Earth plus 325ft (=99.06m) so c is 6,371,000 + 99.06 = 6,371,099.06m. The only unknown in the cosine rule above is b, and we can solve for b to find two mathematical solutions. These are 6371km and 198.12m.
Clearly it is the smaller value we want here! So b = 198.12m.
Now we need angle B, the angle between lines a and c. We get this from the sine rule: sin(B)/b = sin(A)/a. So:
sin(B)=sin(60°)*198.12/6371000 = 2.693*10-5
Angle B = sin-1(2.693*10-5) or B = 2.693*10-5 radians (when an angle is very small its value in radians is approximately equal to its sine). The distance from the ship to the lighthouse along the curve of the Earth is just B*a where B is in radians. So $${\mathtt{distance}} = {\mathtt{2.693}}{\mathtt{\,\times\,}}{{\mathtt{10}}}^{\left(-{\mathtt{5}}\right)}{\mathtt{\,\times\,}}{\mathtt{6\,371\,000}} = {\mathtt{distance}} = {\mathtt{171.571\: \!03}}$$ metres
This is 562.9ft to the nearest tenth of a foot.
Alternatively
Since the distances are relatively small we could assume the surface of the Earth to be flat near the lighthouse, in which case the distance of the ship from the lighthouse is given by 325*tan(60°)ft or
$${\mathtt{distance}} = {\mathtt{325}}{\mathtt{\,\times\,}}\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}{\left({\mathtt{60}}^\circ\right)} = {\mathtt{distance}} = {\mathtt{562.916\: \!512\: \!459\: \!925}}$$ feet
or 562.9ft.
