Solve the triangle. Round lengths to the nearest tenth and angle measures to the nearest degree. a = 7, b = 7, c = 5 A. A = 70°, B = 70°, C = 40° B. A = 69°, B = 69°, C = 42° C. A = 42°, B = 69°, C = 69° D. A = 69°, B = 42°, C = 69°
@Psymon
So you would need to use the law of cosines for this problem. Depending on the angle you want to solve for, you can set up the law of cosines formula differently. Let's just solve for angle A first and let's see if we can set up and solve.
\[a ^{2} = b ^{2} + c ^{2} - 2bc(cosA)\] The sole variable on the left always is the same letter as the cosine term. The cosine term is the angle we want to solve for. So if you plug in the side values for little a, b, c, you think you could isolate cosA?
49=49+25-70.. right so far?
Just dont forget that the -70 is being multiplied by cosA :P good so far.
okay so then it would be.. 49=49+25-490
49=-416?
\[49 = 49+25 - 70(cosA) \] \[49 = 74 - 70(cosA)\] \[-25=-70(cosA)\] \[\frac{ -25 }{ -70 } = cosA\] \[\frac{ 5 }{ 14 } = cosA \] you can't just make the cosA disappear :P
I'm so confused. I thought CosA was just seven?
Okay, there's a difference between a, A, b, B, c, C. The lower case letters are sides and the upper case letters are angles. So I plugged ina, b, c, but I have no idea what CAPITAL A is. Capital letters are always angles, lower case letters are always sides.
So can I just plug in one of the choices then?
Well, we left off with (5/14) = cosA. But we need to solve for A, not just cos A. Do you know how to solve for the angle?
adj and hyp?
We need inverse cosine.
I don't know how to work that out.
It's calculator work to be honest.
Okay, thank you for your help.
Yep, yep :P And ifyou need the other angles, do you remember the law of sines?
I do.
Okay, cool. Good luck then and let's us know again if you need help ^_^
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