QuestionFind the distance between VLF transmitters $P$ and $Q$ given $PB = 180$ ft, $PS = 300$ ft, and $QS = 260$ ft. Options: A. 100 B. 230.

Studdy Solution

STEP 1

Assumptions1. The distance between transmitter and the buoy B is180 feet. . The straight-line distance between the submarine and transmitter is300 feet.
3. The straight-line distance between the submarine and transmitter Q is260 feet.
4. The buoy B, the submarine, and the two transmitters and Q are all located on the same straight line.
5. The submarine is located directly below the buoy B.

STEP 2

We can use the Pythagorean theorem to find the distance from transmitter to the submarine, which is the hypotenuse of a right triangle with sides of180 feet (the distance from to B) and the depth of the submarine (which we'll call d).
$d = \sqrt{PS^2 - PB^2}$

STEP 3

Plug in the given values for PS and PB to calculate the depth of the submarine.
$d = \sqrt{300^2 -180^2}$

STEP 4

Calculate the depth of the submarine.
$d = \sqrt{300^2 -180^2} = \sqrt{90,000 -32,400} = \sqrt{57,600} =240\, feet$

STEP 5

Now, we can use the Pythagorean theorem again to find the distance from transmitter Q to the buoy B, which is the other leg of a right triangle with hypotenuse of260 feet (the distance from Q to) and the depth of the submarine (which we just found to be240 feet).
$QB = \sqrt{QS^2 - d^2}$

STEP 6

Plug in the given values for QS and d to calculate the distance from Q to B.
$QB = \sqrt{260^2 -240^2}$

STEP 7

Calculate the distance from Q to B.
$QB = \sqrt{260^2 -240^2} = \sqrt{67,600 -57,600} = \sqrt{10,000} =100\, feet$

STEP 8

Now that we have the distances from to B and from Q to B, we can add these together to find the total distance from to Q.
$Q = PB + QB$

STEP 9

Plug in the values for PB and QB to calculate the total distance from to Q.
$Q =180 +100$

STEP 10

Calculate the total distance from to Q.
$Q =180 +100 =280\, feet$ The distance between the two VLF transmitters is closest to280 feet.

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QuestionFind the distance between VLF transmitters $P$ and $Q$ given $PB = 180$ ft, $PS = 300$ ft, and $QS = 260$ ft. Options: A. 100 B. 230.

Studdy Solution

STEP 1

STEP 2

We can use the Pythagorean theorem to find the distance from transmitter to the submarine, which is the hypotenuse of a right triangle with sides of180 feet (the distance from to B) and the depth of the submarine (which we'll call d).
$d = \sqrt{PS^2 - PB^2}$

STEP 3

Plug in the given values for PS and PB to calculate the depth of the submarine.
$d = \sqrt{300^2 -180^2}$

STEP 4

Calculate the depth of the submarine.
$d = \sqrt{300^2 -180^2} = \sqrt{90,000 -32,400} = \sqrt{57,600} =240\, feet$

STEP 5

Now, we can use the Pythagorean theorem again to find the distance from transmitter Q to the buoy B, which is the other leg of a right triangle with hypotenuse of260 feet (the distance from Q to) and the depth of the submarine (which we just found to be240 feet).
$QB = \sqrt{QS^2 - d^2}$

STEP 6

Plug in the given values for QS and d to calculate the distance from Q to B.
$QB = \sqrt{260^2 -240^2}$

STEP 7

Calculate the distance from Q to B.
$QB = \sqrt{260^2 -240^2} = \sqrt{67,600 -57,600} = \sqrt{10,000} =100\, feet$

STEP 8

Now that we have the distances from to B and from Q to B, we can add these together to find the total distance from to Q.
$Q = PB + QB$

STEP 9

Plug in the values for PB and QB to calculate the total distance from to Q.
$Q =180 +100$

STEP 10

Calculate the total distance from to Q.
$Q =180 +100 =280\, feet$ The distance between the two VLF transmitters is closest to280 feet.

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QuestionFind the distance between VLF transmitters PPP and QQQ given PB=180PB = 180PB=180 ft, PS=300PS = 300PS=300 ft, and QS=260QS = 260QS=260 ft. Options: A. 100 B. 230.

Studdy Solution

STEP 1

STEP 2

We can use the Pythagorean theorem to find the distance from transmitter to the submarine, which is the hypotenuse of a right triangle with sides of180 feet (the distance from to B) and the depth of the submarine (which we'll call d).d=PS2−PB2d = \sqrt{PS^2 - PB^2}d=PS2−PB2​

STEP 3

Plug in the given values for PS and PB to calculate the depth of the submarine.d=3002−1802d = \sqrt{300^2 -180^2}d=3002−1802​

STEP 4

Calculate the depth of the submarine.d=3002−1802=90,000−32,400=57,600=240 feetd = \sqrt{300^2 -180^2} = \sqrt{90,000 -32,400} = \sqrt{57,600} =240\, feetd=3002−1802​=90,000−32,400​=57,600​=240feet

STEP 5

STEP 6

Plug in the given values for QS and d to calculate the distance from Q to B.QB=2602−2402QB = \sqrt{260^2 -240^2}QB=2602−2402​

STEP 7

Calculate the distance from Q to B.QB=2602−2402=67,600−57,600=10,000=100 feetQB = \sqrt{260^2 -240^2} = \sqrt{67,600 -57,600} = \sqrt{10,000} =100\, feetQB=2602−2402​=67,600−57,600​=10,000​=100feet

STEP 8

Now that we have the distances from to B and from Q to B, we can add these together to find the total distance from to Q.Q=PB+QBQ = PB + QBQ=PB+QB

STEP 9

Plug in the values for PB and QB to calculate the total distance from to Q.Q=180+100Q =180 +100Q=180+100

STEP 10

Calculate the total distance from to Q.Q=180+100=280 feetQ =180 +100 =280\, feetQ=180+100=280feetThe distance between the two VLF transmitters is closest to280 feet.

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QuestionFind the distance between VLF transmitters $P$ and $Q$ given $PB = 180$ ft, $PS = 300$ ft, and $QS = 260$ ft. Options: A. 100 B. 230.

We can use the Pythagorean theorem to find the distance from transmitter to the submarine, which is the hypotenuse of a right triangle with sides of180 feet (the distance from to B) and the depth of the submarine (which we'll call d).
$d = \sqrt{PS^2 - PB^2}$

Plug in the given values for PS and PB to calculate the depth of the submarine.
$d = \sqrt{300^2 -180^2}$

Calculate the depth of the submarine.
$d = \sqrt{300^2 -180^2} = \sqrt{90,000 -32,400} = \sqrt{57,600} =240\, feet$

Plug in the given values for QS and d to calculate the distance from Q to B.
$QB = \sqrt{260^2 -240^2}$

Calculate the distance from Q to B.
$QB = \sqrt{260^2 -240^2} = \sqrt{67,600 -57,600} = \sqrt{10,000} =100\, feet$

Now that we have the distances from to B and from Q to B, we can add these together to find the total distance from to Q.
$Q = PB + QB$

Plug in the values for PB and QB to calculate the total distance from to Q.
$Q =180 +100$

Calculate the total distance from to Q.
$Q =180 +100 =280\, feet$ The distance between the two VLF transmitters is closest to280 feet.