Q: Trigonometry Bearings: Find distance between 2 points based on separate bearings
$begingroup$
Just finished a trig exam with the following problem and most everyone in the class arrived at a very different answer than me. The Professor and most of the class determined that to solve for side AB you would simply use the law of sines:
Problem:
A land developer wants to find the distance across a small lake in the middle of his proposed development. The bearing from A to B is N15°W. The developer leaves point A and travels 66 yards perpendicular to AB to point C. The bearing from C to point B is N75°W. Determine the distance, AB, across the small lake. Round distance to nearest yard.
Class Solution:
(using law of sines)
step1: 66 yards / sin15° = AB / sin75°
step2: (66)(sin75°) / sin15° = AB
step3: 246.32 ≈ AB
My Confusion with this is that the angles are not directly stated in the problem, at least from my understanding of trig bearings. It seems like the Professor and the students assumed that (angle B = 15°) and since (angle A = 90°) then (angle C = 75°).
So I created a little diagram using actual measured angles to figure out what the triangle looked like based on my understanding of bearings (for N15°W = start at North or 90° on a unit circle and move counter clockwise 15°) and from my diagram I determined that (angle C = 30°), (angle A = 90°), therefore (angle B = 60°). I ran through this problem multiple times and even looked at similar problems both in the book and online, and basically want to validate which answer is correct. Thanks for playing!
My solution:
(Using law of sines)
step1: 66 yards / sin60° = AB / sin30°
step2: (66)(sin30°) / sin60° = AB
step3: 38.11 ≈ AB
geometry trigonometry
edited Apr 27 '17 at 17:51
amd
30.4k21050
asked Apr 27 '17 at 15:02
escapetherobotsescapetherobots
1113
$endgroup$
add a comment |
$begingroup$
Just finished a trig exam with the following problem and most everyone in the class arrived at a very different answer than me. The Professor and most of the class determined that to solve for side AB you would simply use the law of sines:
Problem:
A land developer wants to find the distance across a small lake in the middle of his proposed development. The bearing from A to B is N15°W. The developer leaves point A and travels 66 yards perpendicular to AB to point C. The bearing from C to point B is N75°W. Determine the distance, AB, across the small lake. Round distance to nearest yard.
Class Solution:
(using law of sines)
step1: 66 yards / sin15° = AB / sin75°
step2: (66)(sin75°) / sin15° = AB
step3: 246.32 ≈ AB
My Confusion with this is that the angles are not directly stated in the problem, at least from my understanding of trig bearings. It seems like the Professor and the students assumed that (angle B = 15°) and since (angle A = 90°) then (angle C = 75°).
So I created a little diagram using actual measured angles to figure out what the triangle looked like based on my understanding of bearings (for N15°W = start at North or 90° on a unit circle and move counter clockwise 15°) and from my diagram I determined that (angle C = 30°), (angle A = 90°), therefore (angle B = 60°). I ran through this problem multiple times and even looked at similar problems both in the book and online, and basically want to validate which answer is correct. Thanks for playing!
My solution:
(Using law of sines)
step1: 66 yards / sin60° = AB / sin30°
step2: (66)(sin30°) / sin60° = AB
step3: 38.11 ≈ AB
geometry trigonometry
edited Apr 27 '17 at 17:51
amd
30.4k21050
asked Apr 27 '17 at 15:02
escapetherobotsescapetherobots
1113
$endgroup$
1
$begingroup$
What not just $AB = 66 tan 30^0$.
$endgroup$
– Mick
Apr 27 '17 at 15:56
$begingroup$
True, based on my solution, AB = (66)(tan30°), I guess my real question is how everyone else in the class derived the angles based on the bearings. Is (∠C = 30°) or (∠C = 75°)
$endgroup$
– escapetherobots
Apr 27 '17 at 16:16
$begingroup$
If the notation for the bearings is that used by the US Army then your answer is correct.
$endgroup$
– Jens
Apr 27 '17 at 17:20
$begingroup$
1) The correct wording of the bold-faced is "The bearing of B from A is $N 15^0 W$". 2) The simplest thing to do is to ask your professor or your classmates to draw their figures. Then, we can verify which version is correct.
$endgroup$
– Mick
Apr 27 '17 at 17:41
$begingroup$
Here is what one of my classmates sent me as an explanation of what they thought were the values of the triangle. Good times. classmate example
$endgroup$
– escapetherobots
Apr 27 '17 at 21:46
add a comment |
$begingroup$
Just finished a trig exam with the following problem and most everyone in the class arrived at a very different answer than me. The Professor and most of the class determined that to solve for side AB you would simply use the law of sines:
Problem:
A land developer wants to find the distance across a small lake in the middle of his proposed development. The bearing from A to B is N15°W. The developer leaves point A and travels 66 yards perpendicular to AB to point C. The bearing from C to point B is N75°W. Determine the distance, AB, across the small lake. Round distance to nearest yard.
Class Solution:
(using law of sines)
step1: 66 yards / sin15° = AB / sin75°
step2: (66)(sin75°) / sin15° = AB
step3: 246.32 ≈ AB
My Confusion with this is that the angles are not directly stated in the problem, at least from my understanding of trig bearings. It seems like the Professor and the students assumed that (angle B = 15°) and since (angle A = 90°) then (angle C = 75°).
So I created a little diagram using actual measured angles to figure out what the triangle looked like based on my understanding of bearings (for N15°W = start at North or 90° on a unit circle and move counter clockwise 15°) and from my diagram I determined that (angle C = 30°), (angle A = 90°), therefore (angle B = 60°). I ran through this problem multiple times and even looked at similar problems both in the book and online, and basically want to validate which answer is correct. Thanks for playing!
My solution:
(Using law of sines)
step1: 66 yards / sin60° = AB / sin30°
step2: (66)(sin30°) / sin60° = AB
step3: 38.11 ≈ AB
geometry trigonometry
edited Apr 27 '17 at 17:51
amd
30.4k21050
asked Apr 27 '17 at 15:02
escapetherobotsescapetherobots
1113
$endgroup$
Just finished a trig exam with the following problem and most everyone in the class arrived at a very different answer than me. The Professor and most of the class determined that to solve for side AB you would simply use the law of sines:
Problem:
A land developer wants to find the distance across a small lake in the middle of his proposed development. The bearing from A to B is N15°W. The developer leaves point A and travels 66 yards perpendicular to AB to point C. The bearing from C to point B is N75°W. Determine the distance, AB, across the small lake. Round distance to nearest yard.
Class Solution:
(using law of sines)
step1: 66 yards / sin15° = AB / sin75°
step2: (66)(sin75°) / sin15° = AB
step3: 246.32 ≈ AB
My Confusion with this is that the angles are not directly stated in the problem, at least from my understanding of trig bearings. It seems like the Professor and the students assumed that (angle B = 15°) and since (angle A = 90°) then (angle C = 75°).
So I created a little diagram using actual measured angles to figure out what the triangle looked like based on my understanding of bearings (for N15°W = start at North or 90° on a unit circle and move counter clockwise 15°) and from my diagram I determined that (angle C = 30°), (angle A = 90°), therefore (angle B = 60°). I ran through this problem multiple times and even looked at similar problems both in the book and online, and basically want to validate which answer is correct. Thanks for playing!
My solution:
(Using law of sines)
step1: 66 yards / sin60° = AB / sin30°
step2: (66)(sin30°) / sin60° = AB
step3: 38.11 ≈ AB
geometry trigonometry
geometry trigonometry
edited Apr 27 '17 at 17:51
amd
30.4k21050
asked Apr 27 '17 at 15:02
escapetherobotsescapetherobots
1113
edited Apr 27 '17 at 17:51
amd
30.4k21050
asked Apr 27 '17 at 15:02
escapetherobotsescapetherobots
1113
edited Apr 27 '17 at 17:51
amd
30.4k21050
edited Apr 27 '17 at 17:51
amd
30.4k21050
edited Apr 27 '17 at 17:51
amd
30.4k21050
30.4k21050
asked Apr 27 '17 at 15:02
escapetherobotsescapetherobots
1113
asked Apr 27 '17 at 15:02
escapetherobotsescapetherobots
1113
asked Apr 27 '17 at 15:02
escapetherobotsescapetherobots
1113
1113
1
$begingroup$
What not just $AB = 66 tan 30^0$.
$endgroup$
– Mick
Apr 27 '17 at 15:56
$begingroup$
True, based on my solution, AB = (66)(tan30°), I guess my real question is how everyone else in the class derived the angles based on the bearings. Is (∠C = 30°) or (∠C = 75°)
$endgroup$
– escapetherobots
Apr 27 '17 at 16:16
$begingroup$
If the notation for the bearings is that used by the US Army then your answer is correct.
$endgroup$
– Jens
Apr 27 '17 at 17:20
$begingroup$
1) The correct wording of the bold-faced is "The bearing of B from A is $N 15^0 W$". 2) The simplest thing to do is to ask your professor or your classmates to draw their figures. Then, we can verify which version is correct.
$endgroup$
– Mick
Apr 27 '17 at 17:41
$begingroup$
Here is what one of my classmates sent me as an explanation of what they thought were the values of the triangle. Good times. classmate example
$endgroup$
– escapetherobots
Apr 27 '17 at 21:46
add a comment |
1
$begingroup$
What not just $AB = 66 tan 30^0$.
$endgroup$
– Mick
Apr 27 '17 at 15:56
$begingroup$
True, based on my solution, AB = (66)(tan30°), I guess my real question is how everyone else in the class derived the angles based on the bearings. Is (∠C = 30°) or (∠C = 75°)
$endgroup$
– escapetherobots
Apr 27 '17 at 16:16
$begingroup$
If the notation for the bearings is that used by the US Army then your answer is correct.
$endgroup$
– Jens
Apr 27 '17 at 17:20
$begingroup$
1) The correct wording of the bold-faced is "The bearing of B from A is $N 15^0 W$". 2) The simplest thing to do is to ask your professor or your classmates to draw their figures. Then, we can verify which version is correct.
$endgroup$
– Mick
Apr 27 '17 at 17:41
$begingroup$
Here is what one of my classmates sent me as an explanation of what they thought were the values of the triangle. Good times. classmate example
$endgroup$
– escapetherobots
Apr 27 '17 at 21:46
1
1
$begingroup$
What not just $AB = 66 tan 30^0$.
$endgroup$
– Mick
Apr 27 '17 at 15:56
$begingroup$
What not just $AB = 66 tan 30^0$.
$endgroup$
– Mick
Apr 27 '17 at 15:56
$begingroup$
True, based on my solution, AB = (66)(tan30°), I guess my real question is how everyone else in the class derived the angles based on the bearings. Is (∠C = 30°) or (∠C = 75°)
$endgroup$
– escapetherobots
Apr 27 '17 at 16:16
$begingroup$
True, based on my solution, AB = (66)(tan30°), I guess my real question is how everyone else in the class derived the angles based on the bearings. Is (∠C = 30°) or (∠C = 75°)
$endgroup$
– escapetherobots
Apr 27 '17 at 16:16
$begingroup$
If the notation for the bearings is that used by the US Army then your answer is correct.
$endgroup$
– Jens
Apr 27 '17 at 17:20
$begingroup$
If the notation for the bearings is that used by the US Army then your answer is correct.
$endgroup$
– Jens
Apr 27 '17 at 17:20
$begingroup$
1) The correct wording of the bold-faced is "The bearing of B from A is $N 15^0 W$". 2) The simplest thing to do is to ask your professor or your classmates to draw their figures. Then, we can verify which version is correct.
$endgroup$
– Mick
Apr 27 '17 at 17:41
$begingroup$
1) The correct wording of the bold-faced is "The bearing of B from A is $N 15^0 W$". 2) The simplest thing to do is to ask your professor or your classmates to draw their figures. Then, we can verify which version is correct.
$endgroup$
– Mick
Apr 27 '17 at 17:41
$begingroup$
Here is what one of my classmates sent me as an explanation of what they thought were the values of the triangle. Good times. classmate example
$endgroup$
– escapetherobots
Apr 27 '17 at 21:46
$begingroup$
Here is what one of my classmates sent me as an explanation of what they thought were the values of the triangle. Good times. classmate example
$endgroup$
– escapetherobots
Apr 27 '17 at 21:46
add a comment |
1 Answer
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I’m going to add a couple of points to your diagram for clarity:
Since $angle{BAC}$ and $angle{EAD}$ are both right angles, then $angle{EAB}=angle{DAC}$. The lines $overline{AD}$ and $overline{EC}$ are parallel, so $angle{EAC}=angle{DAC}=15°$. Finally, $overline{EC}perpoverline{DC}$, so $angle{ECB}=90°-75°=15°$.
answered Apr 27 '17 at 18:11
amdamd
30.4k21050
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$begingroup$
I’m going to add a couple of points to your diagram for clarity:
Since $angle{BAC}$ and $angle{EAD}$ are both right angles, then $angle{EAB}=angle{DAC}$. The lines $overline{AD}$ and $overline{EC}$ are parallel, so $angle{EAC}=angle{DAC}=15°$. Finally, $overline{EC}perpoverline{DC}$, so $angle{ECB}=90°-75°=15°$.
answered Apr 27 '17 at 18:11
amdamd
30.4k21050
$endgroup$
add a comment |
$begingroup$
I’m going to add a couple of points to your diagram for clarity:
Since $angle{BAC}$ and $angle{EAD}$ are both right angles, then $angle{EAB}=angle{DAC}$. The lines $overline{AD}$ and $overline{EC}$ are parallel, so $angle{EAC}=angle{DAC}=15°$. Finally, $overline{EC}perpoverline{DC}$, so $angle{ECB}=90°-75°=15°$.
answered Apr 27 '17 at 18:11
amdamd
30.4k21050
$endgroup$
add a comment |
$begingroup$
I’m going to add a couple of points to your diagram for clarity:
Since $angle{BAC}$ and $angle{EAD}$ are both right angles, then $angle{EAB}=angle{DAC}$. The lines $overline{AD}$ and $overline{EC}$ are parallel, so $angle{EAC}=angle{DAC}=15°$. Finally, $overline{EC}perpoverline{DC}$, so $angle{ECB}=90°-75°=15°$.
answered Apr 27 '17 at 18:11
amdamd
30.4k21050
$endgroup$
I’m going to add a couple of points to your diagram for clarity:
Since $angle{BAC}$ and $angle{EAD}$ are both right angles, then $angle{EAB}=angle{DAC}$. The lines $overline{AD}$ and $overline{EC}$ are parallel, so $angle{EAC}=angle{DAC}=15°$. Finally, $overline{EC}perpoverline{DC}$, so $angle{ECB}=90°-75°=15°$.
answered Apr 27 '17 at 18:11
amdamd
30.4k21050
answered Apr 27 '17 at 18:11
amdamd
30.4k21050
answered Apr 27 '17 at 18:11
amdamd
30.4k21050
answered Apr 27 '17 at 18:11
amdamd
30.4k21050
30.4k21050
add a comment |
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1
$begingroup$
What not just $AB = 66 tan 30^0$.
$endgroup$
– Mick
Apr 27 '17 at 15:56
$begingroup$
True, based on my solution, AB = (66)(tan30°), I guess my real question is how everyone else in the class derived the angles based on the bearings. Is (∠C = 30°) or (∠C = 75°)
$endgroup$
– escapetherobots
Apr 27 '17 at 16:16
$begingroup$
If the notation for the bearings is that used by the US Army then your answer is correct.
$endgroup$
– Jens
Apr 27 '17 at 17:20
$begingroup$
1) The correct wording of the bold-faced is "The bearing of B from A is $N 15^0 W$". 2) The simplest thing to do is to ask your professor or your classmates to draw their figures. Then, we can verify which version is correct.
$endgroup$
– Mick
Apr 27 '17 at 17:41
$begingroup$
Here is what one of my classmates sent me as an explanation of what they thought were the values of the triangle. Good times. classmate example
$endgroup$
– escapetherobots
Apr 27 '17 at 21:46