Three known sides need angles

**Jarod_C** · April 9th 2010, 08:55 AM

a=152.4cm b=same c=118.11cm Now I need the angles.

I'm building a chicken coop and can't figure out the the angles. I guess two semesters of algebra and trig two years ago doesn't do me any good now if I don't remember any of it. I think the problem is not remembering my calculator functions. I got 0.3875 for A, and I think I just need to hit COS but when I do it doesn't sound right. Maybe I'm way off. Anyway if someone could show their work I'd like to know where I went wrong.

Thanks alot for the help.

Edit:I'll try to show my work below to see where I went wrong.

$(b^2+c^2-a^2)/2bc=\cos A$
,so
$(13949.9721)/35999.928=0.3875$

**qmech** · April 9th 2010, 09:11 AM

The law of cosines will help.

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

a,b,c are the sides of the triangle. From there you can get cos(theta) and from there get theta.

**Soroban** · April 9th 2010, 09:39 AM

Hello, Jarod_C!

Your work is correct . . . take it a step further.

$a \:=\: b\:=\:152.4\text{ cm,}\;c\:=\:118.11\text{ cm}$
Find angles $A,B,C.$

I'll try to show my work below to see where I went wrong.

$\cos A \:=\:\frac{b^2+c^2-a^2}{2bc}$

So: . $\cos A \:=\:\frac{13949.9721}{35999.928} \:=\:0.3875$

Then: . $A \;=\;\cos^{-1}(0.3875) \:=\:67.20096873^o \quad\Rightarrow\quad \boxed{A \;\approx\;67.2^o}$

The triangle is isosceles, so: . $\boxed{B \:=\:67.2^o}$

Hence: . $C \;=\;180^o - 67.2^o - 67.2^o \quad\Rightarrow\quad \boxed{C\:=\:45.6^o}$

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Check

$\cos C \:=\:\frac{152.4^2 + 152.4^2 - 118.11^2}{2(152.4)^2} \;=\;0.6996875$

Therefore: . $C \:=\:45.59806254^o \quad\Rightarrow\quad \boxed{C \;\approx\; 45.6^o}$

**Jarod_C** · April 9th 2010, 10:52 AM

Tyvm, that's what I was needing. I'm still stumped on my calculator though. What function changes decimal from 0.3875 to $67.2^o$

I think I'm going to lay off the ps3 because I used to be able to do this in my sleep.

Edit:LOL I'm rusty at this. I got my function by reading the problem. 03875 cos-1 = 67.2 deg. Thanks it really helped.

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