# Trigonometric Word Problem involving Law of Sines & Law of Cosines; I am so clueless.

• November 20th 2009, 05:29 AM
mdebourg
Trigonometric Word Problem involving Law of Sines & Law of Cosines; I am so clueless.
Hi there, I just heard about this website from a math tutor after she couldn't help me figure out these problems herself! Hopefully someone here from around the world could help me with these two problems...

1. A rocket tracking station has two telescopes A and B, placed 1.4 miles apart. The telescopes lock onto a rocket and transmit their angles of elevation to a computer after a rocket launch. What is the distance to the rocket from telescope B at the moment when both tracking stations are directly east of the rocket telescope A reports an angle of elevation of 29 degrees and telescope B reports an angle of elevation of 49 degrees?

I keep doing this problem over and over, and keep getting the wrong answer, my only resolution would be that maybe I'm setting the problem up wrong..?

2. A famous golfer tees off on a straight 380 yard par 4 and slices his drive to the right. The drive goes 280 yards from the tee. Using a 7-iron on his second shot, he hits the ball 160 yards and it lands inches from the hole. How many degrees (to the nearest degree) to the right of the line from the tee to the hole, did he slice his drive?

For this one I don't even know where to start...(Worried)

If anyone would take the time to try and solve any of these it would be tremendously appreciated!
• November 20th 2009, 06:05 AM
masters
Quote:

Originally Posted by mdebourg
Hi there, I just heard about this website from a math tutor after she couldn't help me figure out these problems herself! Hopefully someone here from around the world could help me with these two problems...

1. A rocket tracking station has two telescopes A and B, placed 1.4 miles apart. The telescopes lock onto a rocket and transmit their angles of elevation to a computer after a rocket launch. What is the distance to the rocket from telescope B at the moment when both tracking stations are directly east of the rocket telescope A reports an angle of elevation of 29 degrees and telescope B reports an angle of elevation of 49 degrees?

I keep doing this problem over and over, and keep getting the wrong answer, my only resolution would be that maybe I'm setting the problem up wrong..?

2. A famous golfer tees off on a straight 380 yard par 4 and slices his drive to the right. The drive goes 280 yards from the tee. Using a 7-iron on his second shot, he hits the ball 160 yards and it lands inches from the hole. How many degrees (to the nearest degree) to the right of the line from the tee to the hole, did he slice his drive?

For this one I don't even know where to start...(Worried)

If anyone would take the time to try and solve any of these it would be tremendously appreciated!

Hi mdebourg,

If I understand #2 correctly, construct a triangle this way.

From T to H is a distance of 380. From T to S is a distance of 280. From S to H is a distance of 160. You need to find the angle between the 380 distance and the 280 distance.

$c^2=a^2+b^2-2ab \cos \theta$

$160^2=380^2+280^2-2(380)(280)\cos \theta$
• November 20th 2009, 08:51 AM
Soroban
Hello, mdebourg!

Quote:

1. A rocket tracking station has two telescopes A and B, placed 1.4 miles apart.
The telescopes lock onto a rocket and transmit their angles of elevation to a computer.
What is the distance to the rocket from telescope B at the moment
when both tracking stations are directly east of the rocket,
and telescope A reports an angle of elevation of 29°
and telescope B reports an angle of elevation of 49° ?

It took me two tries to get the diagram . . .

Code:

    R *       : *  *       :  *  20°  *       :    *          *       :      *              *       :        *                  *       :      49° *  131°          29°  *       * - - - - - - * - - - - - - - - - - - - - *       W            B          1.4            A

Since $\angle RBW = 49^o:\;\;\angle RBA = 131^o$

Then in $\Delta RBA\!:\;\angle ARB = 20^o$

Law of Sines: . $\frac{BR}{\sin29^o} \:=\:\frac{1.4}{\sin20^o}$

Got it?

• November 21st 2009, 05:51 AM
mdebourg
Thank you Masters and Soroban for such amazing help! I completely understand these types of problems now, super appreciated, super helpful thanks a lot!!!!(Clapping)