How Do I Solve Trigonometry Word Problems for My Exam?

[aturman0803]

I'm having a really hard time solving these word problems. ANY info on how to get the pictures drawn, which sides or angles to solve for, and instructions on how to solve would be greatly appreciated!

1. A tv antenna on top of a house is 4.50 ft tall. From a point on level ground the angles of elevation to the top and base of the antenna are 59.5deg and56.3deg respectively. How tall is the house?
--I drew a house with a antenna on top and labeled the antenna side 4.50 ft, not sure if I drew the angles of elevation correct...I have them both coming from the SAME point on the ground--one to the top of antenna, one to the bottom of antenna--and labeled those angles. So I have a ASA triangle. Now how do I find the height of the house?

2. Two boats leave a dock together, traveling in courses that are straight lines with a difference of 47deg40min btwn their courses. One boat travels 30.2 kph, the other 42.6 kph. How far apart are the boats after 2 hours?
--On this one I have a picture I think is right and found the side opposite of the angle given. Do I have to use the 2 hours part for anything or is that just there to throw me off? I used the law of cosines and got 31.5km as my answer. Is this correct?

3. Courtney can row her boat at 10mph in still water. She rows across a river at a bearing of 140degrees. The current of the river coming from the west has a speed of 5kph. Find the resulting speed of the boat and its resulting direction of travel.
--On this one I drew my bearing of 140 and on that line labeled it 10kph. I then drew a vector going from west to east to represent the wind, with the arrow of said vector lining up with arrow (and also labeled it 5 kph) of original 140deg bearing. Now what exactly am I supposed to solve for?

4. A plane is flying at airspeed of 315 mph on a bearing of 215.0 degrees. A 22.4 mph wind is blowing north. What is the true course (bearing) of said plane and what is its groundspeed?
--I have drawn the picture i have a side 315, 22.4, and x. Is that right, and if so, how do I solve?

Word problems where I have to set up the picture are definitely not my strong suit. Somebody, anybody--help--PLEASE! THANK YOU! :)

 

[AdamX1980X]

For #1 draw a house from a side then draw the antenna with the measurement of how high above the roof it is. Now draw two lines, one from the bottom of the antenna to the ground and another from the top of the antenna to the ground. Look at your picture ask yourself which line would have the greater angle between the level ground and the lines that we have drawn. Draw a line from the center of the house to the antenna. This should form a right triangle. Now the height would be a sine function. So you should have sine of an angle = to a distance call it x and another sin of an angle = x+4.5. Hope this helps with the first one. Now after setting this up I subtracted one angle from the other got 3.2 degrees. set the sin of that = to the height of the antenna. and set the sin of the 56.3 degrees = x. solve for x using a proportionality. I think this is a correct way to solve this

 

[aturman0803]

AdamX1980X,
Thank you, I will try that and see if I can get it. The only thing is, with my picture, I am seeing no way to draw a line from the house to the antenna. Maybe I have my picture drawn all wrong.

On a different note: I think I have solved the boat problem. I doubled the side lengths because of the 2 hours and got 60.4 and 85.2 for the new side lengths. I then used the law of cosines and my answer is 63.1 km apart. Can anyone confirm that this is correct?

 

[JThompson]

AdamX1980X,
Thank you, I will try that and see if I can get it. The only thing is, with my picture, I am seeing no way to draw a line from the house to the antenna. Maybe I have my picture drawn all wrong.

I'm not sure what you mean by this. If you let h be the height of the house, then the whole thing should be h+4.5 . This should be a side of the triangle with a 59.5 degree interior angle opposite this side. Then you'll have another triangle with an interior angle of 56.3 degrees with an opposite side of length h.

If drawn correctly, you should get two right triangles which you should be able to use the law of sines on to find h.