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The problem is in the first picture and the questions are in the second one. I have no idea how to do any of this.

The problem is in the first picture and the questions are in the second one. I have-example-1
The problem is in the first picture and the questions are in the second one. I have-example-1
The problem is in the first picture and the questions are in the second one. I have-example-2
User Tryp
by
8.1k points

1 Answer

3 votes
Problem A
You are given that CB = 271 feet. You are also given that <CAB = 30 degrees.

So you can use Sin(<CAB) = opposite / hypotenuse or
hypotenuse = opposite / sin(<CAB)
hypotenuse = 271 / 0.5
hypotenuse (AB) = 542 feet.

Problem B
CD = 772 feet Given in table
<ACD = 74o Given on drawing.

We need to find AC which is not given
Cos 30 = adjacent / hypotenuse.
adjacent = hypotenuse * Cos(30)
adjacent = AB * Cos(30)
Adjacent = 542 * cos(30)
Adjacent = AC = 469.4 feet

Now we need to start again
AC = 469.4
<ACD = 74
CD = 772
AD = ???

AD^2 = AC^2 + CD^2 - 2 * AC * CD * Cos(74)
AD^2 = 595984 + 220336.4 - 399538
AD^2 = 416782.1
AD = sqrt(416782.1)
AD = 645.59

Problem 3
I'll set it up for you. And give you the answer I get, but I'm going to leave it to you mostly. It just follows what I've done above.

DC = 645.59
CE = 561
DE = 543

DC^2 = CE^2 + DE^2 - 2* CE * DE * Cos(E)
645.59^2 = 561^2 + 543^2 - 2*561*543 * Cos(E)
Cos(E) = -0.3164
E = cos-1(0.3164)
E = 108.45 degrees.





User Curlyreggie
by
8.3k points

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