Find Distance From Ship To Shore

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This question is located in my precalculus textbook chapter SOLVING RIGHT TRIANGLES. However, can I apply The Law of Sines to find the distance from the ship to shore? If so, how do I set it up?

Question:

The navigator of a ship at sea spots two lighthouses that she sees as ones 3 miles apart along a straight seashore. She determines that the angles formed between the two line-of-sight observations of the lighthouses and the line from the ship directly to shore are 15 degrees and 35 degrees.

How far is the ship from shore?
 
"Law of Sines" Why?

How about:

S is the ship
A is one lighthouse
B is the other lighthouse {Note: I drew B on the Left of A}
D is the point between them, but the point on shore closest to the ship.

angle(SDA) = angle(SDB) = 90º

Then,

angle(DSA) = 35º
angle(DSB) = 15º
length(BA) = 3 miles

The tangent function suggests:

tan(35º) = length(DA) / length(DS)
tan(15º) = length(DB) / length(DS)

Also, length(DA) + length(DB) = length(BA) = 3 miles

You should be able to solve that for length(DS).

HOWEVER

How about:

S is the ship
A is one lighthouse
B is the other lighthouse {Note: I drew D on the Right of A}
D is a point NOT between them, but still the point on shore closest to the ship.

angle(SDA) = angle(SDB) = 90º

Then,

angle(DSA) = 15º
angle(DSB) = 35º
length(BA) = 3 miles

The tangent function suggests:

tan(15º) = length(DA) / length(DS)
tan(35º) = length(DB) / length(DS)

Also, length(DB) = length(DA) + length(AB) = length(DA) + 3 miles

You should be able to solve that for length(DS).

This problem statement shuold be rewritten to clarify intent.
 
One More Time

If distance AB = 3 miles, then distance DB = 1.5 miles = distance DA, right?

If so, I can then use your tangent function set up to find the distance from the ship to shore., right?

Interval
 
It's tough to answer, since I gave two scenarios. It is likely I could figure out which you are using, but part of mathematics is being clear, so I'll make you tell me.

Note: I did not invent the tangent function. :D
 
interval said:
This question is located in my precalculus textbook chapter SOLVING RIGHT TRIANGLES. However, can I apply The Law of Sines to find the distance from the ship to shore? If so, how do I set it up?

Question:

The navigator of a ship at sea spots two lighthouses that she sees as ones 3 miles apart along a straight seashore. She determines that the angles formed between the two line-of-sight observations of the lighthouses and the line from the ship directly to shore are 15 degrees and 35 degrees.

How far is the ship from shore
Click to expand...
1--Draw vertical line AC, A at the top and horizontal line CB, B to the left of C and perpendicular to AC.
2--Draw line AB.
3--Draw line EB, point E on AC.
4--Point B represents the ship, point A represents the farthest lighthouse and point E the nearest lighthouse,
5--Angle EBC = 15º and angle ABC = 35º making angle EBA 20º.
6--BC represents the distance of the ship from the shore.
7--Angle BEWC = 75º ,akong angle BEA 105º and angle BAC = 55º.
8--Then, 3/sin20 = BA/sin105 making BA = 8.472 miles.
9--Therefore, BC = 8.472sin55 = 6.940 miles.
 
tkhunny said:
How about:

S is the ship
A is one lighthouse
B is the other lighthouse {Note: I drew B on the Left of A}
D is the point between them, but the point on shore closest to the ship.
Click to expand...
SD = the distance we seek; let SD = h and DA = x
Then tan35 = x/h and tan15 = (3 - x)/h.

Solving for h and equating yields x -- 2.16953

tan35 = 2.16953/h yieldimg h = 3.09841 = SD
 
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