The obscure triangle has side lengths of 16 in, 13 in, and 20 in. I am assuming side a is 13, side b is 16, and side c is 20. We are looking for angle x between side a and side b. I have tried to use the formula for cos(C) because angle C is supposed to be between side b and side a. However, it is not coming out to any of the answer choices. My work is down below.

Question

cos(C)=(a^2+b^2-c^2)/(2ab)
=(13^2+16^2-20^2)/(2*13*16)
=(169+256-400)/(416)
=25/416
=0.06009615384
cos(C)=3.44325598 degrees

The answer choices are:
a. 14
b. 40
c. 53
d. 86

I also tried cos(A) and cos(B) just to see if I was wrong, but they also did not come out to any of the answer choices.

cos(A)=(b^2+c^2-a^2)/(2bc)
=(16^2+20^2-13^2)/(2*16*20)
=(256+400-169)/(640)
=487/640
=0.7609375
cos(A)=43.6 degrees

cos(B)=(a^2+c^2-b^2)/(2ac)
=(13^2+20^2-16^2)/(2*13*20)
=(169+400-256)/(2*13*20)
=313/520
=0.60192307692
cos(B)=34.48765.... degrees

I am completely and utterly confused. Thank you!

Answer 1

I don't know how you did this step:
cos(C)=0.06009615384
cos(C)=3.44325598 degrees

It should immediately have raised a "what?" flag. You know that cos(90)=0, so C should be close to 90. You did C=arcsin(0.06009)
So, you not only pushed the wring button, you mislabeled it as cos(C) rather than C.

Answer 2

Hi Steve. Sorry, but I have read over what you said a couple times but I am still confused. How did I mislabel it as cos(C) rather than C?

Answer 3

You typed:

cos(C)=0.06009615384
cos(C)=3.44325598 degrees

You should have typed
I don't know how you did this step:
cos(C)=0.06009615384
C=3.44325598 degrees BUT still wrong, that is sin, not cos
cos^-1 (0.06) = about 86.6 degrees

Answer 4

Law of Cosines Practice:
1. (D): 87
2. (B): 16.9
3. (A): 40.0 degrees
4. (B): 25.4 ft.
Law of Cosines Quick Check:
1. (A): 1.7
2. (D): 64.4
3. (A, E, F): 140, 24, & 16(in order.)
4. (C): 106 mi.