Trigonometry Word Problem ... Question 3

**shadow6** · June 12th 2009, 12:16 PM

The following questions are for a final class project. Obviously these questions are very hard, and I am unsure how to completely answer these with the proper and complete work shown. Help on answering these questions with work/ steps would be greatly appreciated.

9. Two airplanes leave the airport at the same time. One travels 355 km/h and the other 450km/h. Two hours later they are 800km apart. Find the angle between their courses.

10. Two helicopters flying at an altitude of 250m are 2000m apart when the spot a life raft below. It is directly between them. The angle of depression (AoD) between one and the raft is 37°. The AoD between the other and the raft is 49°. Both are flying 170km/h. How long to the nearest second will it take the nearest aircraft to reach the raft?

**Shyam** · June 12th 2009, 05:56 PM

Originally Posted by shadow6

9. Two airplanes leave the airport at the same time. One travels 355 km/h and the other 450km/h. Two hours later they are 800km apart. Find the angle between their courses.

Please see the diagram,

In 2 hours, the distance travelled by plane 1 is = 2 (335) = 710 km = PA

In 2 hours, the distance travelled by plane 2 is = 2 (450) = 900 km = PB

AB = 800 km,

Use cos-law to find angle APB,

$\cos \angle APB = \frac{710^2+900^2-800^2}{2\times 710 \times 900}<br />$

$<br /> \angle APB = 58^\circ <br />$

**aidan** · June 12th 2009, 07:35 PM

Originally Posted by shadow6

The following questions are for a final class project. Obviously these questions are very hard, and I am unsure how to completely answer these with the proper and complete work shown. Help on answering these questions with work/ steps would be greatly appreciated.

9. Two airplanes leave the airport at the same time. One travels 355 km/h and the other 450km/h. Two hours later they are 800km apart. Find the angle between their courses.

10. Two helicopters flying at an altitude of 250m are 2000m apart when the spot a life raft below. It is directly between them. The angle of depression (AoD) between one and the raft is 37°. The AoD between the other and the raft is 49°. Both are flying 170km/h. How long to the nearest second will it take the nearest aircraft to reach the raft?

For 9.
You have a triangle with all three sides given:
(355 km/hr)(2hr) = 710 km.
(450 km/hr)(2hr) = 900 km.
distance apart = 800 km.

Use the cosine law

----------------------------------------------------------
For 10. You have invalid data. The 250m altitude is wrong, or the Angles of Depression are wrong, or the raft is NOT between the two aircraft, or you have two rafts and the pilots do not know it. Or there could be other pieces that make this non solvable.

**shadow6** · June 13th 2009, 05:59 AM

TYVM for the help.

These questions are starting to make a lot more sense to me now.

Aidan, I had a friend copy these questions down on paper. The writing wasn't that great but I was able to make it out. When I get the real paper again, I'll see if any typos were made.

**shadow6** · June 15th 2009, 03:51 PM

"For 10. You have invalid data. The 250m altitude is wrong, or the Angles of Depression are wrong, or the raft is NOT between the two aircraft, or you have two rafts and the pilots do not know it. Or there could be other pieces that make this non solvable."

I overlooked this question with my teacher, and it is solvable. My teacher said that the "250m altitude" was unneeded to solve the question, though everything else in the question was worded correctly. Some people have solved it already (not sure if they're right though), though I am very confused in what to do. Any suggestions to start off with?

**shadow6** · June 15th 2009, 05:30 PM

Ok I attempted a diagram to display this situation. Now I need to use the 170km/h to convert measurement or something... Please help...

**aidan** · June 15th 2009, 06:00 PM

Originally Posted by shadow6

Ok I attempted a diagram to display this situation. Now I need to use the 170km/h to convert measurement or something... Please help...

The horizontal distance from point A to the raft is ALTITUDE $\times \cot (37deg)$

The horizontal distance from point B to the raft is ALTITUDE $\times \cot (49deg)$

170km/hr converted

$\frac {170km \times 1000m/km}{1 hr \times 3600 seconds per hour}$ = 47.222 m/sec

**shadow6** · June 15th 2009, 06:01 PM

If my diagram was right, then my final answer was this:

It will take aircraft B approximately 43 minutes to reach the life raft.

Correct me if I'm wrong, Thanks.

**shadow6** · June 15th 2009, 06:05 PM

My conversion was like this:

1km = 1000m
170km = 170000m

1206.5m / 170000m = 0.0071 hours.

0.0071 * 60 = 0.426mins (approx 0.43)

0.43 * 60 = 25.8 seconds

Correction final answer:

It will take aircraft B approximately 26 seconds to reach the raft.

**Shyam** · June 15th 2009, 06:11 PM

Originally Posted by shadow6

Ok I attempted a diagram to display this situation. Now I need to use the 170km/h to convert measurement or something... Please help...

Your diagram is RIGHT. The altitude 250 is irrelevant to the situation.

$\angle \;C = 180 -37 - 49 = 94^\circ$

Use SINE LAW to find distances AC and BC in your diagram.

$\frac{a}{\sin A}=\frac{b}{\sin B}= \frac{c}{\sin C}$

$\frac{a}{\sin 37}=\frac{b}{\sin 49}= \frac{2000}{\sin 94}$

$\Rightarrow \frac{a}{\sin 37}= \frac{2000}{\sin 94}$

$a= \frac{2000\; \sin 37}{\sin 94} = 1206.569 \;m = BC$

Similarly,

$\Rightarrow \frac{b}{\sin 49}= \frac{2000}{\sin 94}$

$b= \frac{2000\; \sin 49}{\sin 94} = 1503.1\;m = AC$

BC distance comes out smaller, that plane at point B is near to life-raft.

Speed of plane = 170 km/h $= \frac{170}{3.6}\;m/s =47.22\;m/s$

Time taken by plane at B to reach life-raft $= \frac{distance}{speed} = \frac{1206.569}{47.22} =25.55\;seconds$

Hope it helps !

**shadow6** · June 15th 2009, 06:19 PM

Thank you both for the help. Sorry for asking so many questions recently. Tomorrows my last day of actual classes till my exam and I am really desperate for clarification on my final assignment questions.

**Shyam** · June 15th 2009, 06:32 PM

Originally Posted by shadow6

Thank you both for the help. Sorry for asking so many questions recently. Tomorrows my last day of actual classes till my exam and I am really desperate for clarification on my final assignment questions.

Please see my last post again. I made some changes at the end.

**shadow6** · June 15th 2009, 06:40 PM

Yes, it did help a lot! Thanks very much again.

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