# Can't figure these 2 out.......angle and distance

• February 26th 2011, 06:58 AM
Can't figure these 2 out.......angle and distance
See the attached form. Problem 3 and 5. I have tried several times and can't seem to figure out how to get the angle in 3 and the distance in 5.

I got 5 with an answer of 5.0176 inches and it's 5.024 inches.
I STILL can't get 3 with all angles I can find and lengths.!!!!!

Thanks for any help given.
Joanne
• February 26th 2011, 08:09 AM
emakarov
For 3, you have $R\sin(90-y)=R\sin(42^\circ50')+1.480$, from where y can be found.

Attachment 20966
For 5, $\tan(57/2)=\tan\angle OAC=OC/AC$, from where AC can be found. (See the attached image above.) Also, $\tan 57=DB/AB$, from where AB can be found. So, we know the horizontal location of A (3.400 - AB) and therefore the location of C.
• February 26th 2011, 09:24 AM
still stuck on the #3, Not getting what your saying. Can you explain a bit more on calculating that please, thanks.
• February 26th 2011, 10:20 AM
emakarov
Are you not sure about the equation I wrote or how to solve it? Let x be the angle between y and $42^\circ50'$. Then $R\sin(x+42^\circ50')$ is the vertical coordinate of the center of the top circle. It is equal to 1.480 plus the vertical coordinate of the center of the rightmost circle. The latter is $R\sin42^\circ50'$. In all this, R = 4.646.

To find y, divide both sides by R and take arcsine of both sides. What I get, though, is $y = 3^\circ13'$, not $3^\circ12'$, as the answer seems to say.
• February 26th 2011, 12:45 PM
You still have not said whether you understand why $R\sin(90-y)=R\sin(42^\circ50')+1.480$ is true. Assuming you do, divide both sides by R = 4.646 to get $\sin(90-y)=\sin(42^\circ50')+1.480/4.646$. Taking $\sin^{-1}$ of both sides gets $90-y=\sin^{-1}(\sin(42^\circ50')+1.480/4.646)$. Since $42^\circ50'=(42+5/6)^\circ$, evaluating the right-hand side gives $86.780^\circ$, i.e., $y=3.220^\circ$. Finally, $0.220\cdot 60\approx 13$, so $3.22^\circ=3^\circ 13'$.