SOLUTION: A helicopter leaves a point P and flies on a bearing of 162 for 3km to a point Q, and then due west to a point R which is on a bearing of 220 from P. Find PR. Find RQ. The flight t

Algebra ->  Trigonometry-basics -> SOLUTION: A helicopter leaves a point P and flies on a bearing of 162 for 3km to a point Q, and then due west to a point R which is on a bearing of 220 from P. Find PR. Find RQ. The flight t      Log On


   

Question 1205527: A helicopter leaves a point P and flies on a bearing of 162 for 3km to a point Q, and then due west to a point R which is on a bearing of 220 from P. Find PR. Find RQ. The flight takes 4 minutes . Find the speed of the helicopter in kilometer per hour
Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
i get:

PR = 3.724548327 kilometers
RQ = 3.321144499 kilometers.

total trip from P to Q to R = 6.321144499 kilometers.
that took 4 minutes = 4/60 hours.
rate * time = distance.
rate * 4/60 = 6.321144499.
solve for rate to get:
rate = 6.321144499 * 60 / 4 = 94.81716749 kilometers per hour.

my diagram is shown below along with my scribbles.



the locations were from P to Q to R.
that would be line segments PQ and RQ.

line segment RQ is composed of line segments RC and CQ.

triangles PQC and PRC were formed.
angle QPC is 18 degrees because it is 180 minus 162.
162 degrees was the bearing from point P to point Q.

angle RPC is 40 degrees because it is 220 minus 180.
220 degrees was the bearing from point P to point R.

i used storage locations to hold the results from the various calculations required.

the length of PQ was given as 3 kilometers.
the length of CQ was based on the sine of 18 degrees and was equal to .92705... kilometers that were stored in SLA.
SLA is storage location A on my calculator.

the length of PC was based on the cosine of 18 degrees and was equal to 2.85316... that were stored in SLB.
SLB is storage location B on my calculator.

since i found the length of PC, i was able to use that to get the lengths of PR and RC.
RC was 2.39409... and stored in SLC.
PR was 3.72454... and stored in SLD.

RQ was the sum of RC and CQ equal to 3.32114... and stored in SLE.

the total route traveled by the helicopter was the length of PQ and QC and RC which was equal to 6.32114... and stored in SLF.

to find the speed of the helicopter, i used the rate * time = distance formula.
the time was 4 minutes = 4/60 hours.
the distance was 6.32114.... which was stored in SLF.
the formual because rate * 4/60 = 6.32114...
i solved for rate to get rate = 94.817... kilometers per hour which was stored in SLG.

hopefully, you can follow this.
if you have any questions, feel free to send me an email through algebra.com.
theo