Algebra II/ Trig
posted by Sha .
Should the triangle be solved beginning with Laws of Sines or Laws of Cosines. Then solve the triangle and round to the nearest tenth. a=16, b=13, c=10
I got this answer but it is wrong and don't understand where I made the mistake.
a=16, b=13, c=10
a^2=b^2+c^22bc Cos A
16^2=13^2+10^22(13)(10) Cos A
16^213^210^2/2(13)(10)=2(13)(10)/2(13)(10)
13/260 Cos A = 0.05=Cos A = 152=A
sinB/b=SinA/a
Sin B/13=Sin152/16
Sin B = 13 Sin 152/16

Whay is A 152deg? Why not 87 deg?

i don't know
can you explain step by step 
Huh?
cosA=.05
A= arccos.05
Use your calculator. Think. When is cosine of an angle near zero? 
I'd guess it's the area is what the question wants so we need only one of the angles.
Now, CosA = 0.05 but A is not 152.
[For one reason Cosine is ve in the 2nd quadrant so the angle must in the 1st quadrant [0<A<90]
Angle A = 87.1 Degrees
So when you have the angle A = 87.1 the area of the triangle is (1/2)*bcSinA
So the area is (0.5)*13*10*Sin(87.1)
Area = 65*Sin(87.1)
Area = 64.92
Rounded to the 1 decimal place then is
Area = 64.9
Hope that helps. 
Ganon, giving answers seldom helps a student, at least in my experience.

The procedure that I did is correct just the math is incorrect...is this what you are telling me?

When all you are given are the three sides, you have no choice,
You HAVE TO use the Cosine Law to find one of the angles.
Then you can use the Sine Law to find the second angle.
After that the third angle is easy.
When using the Sine Law to find an angle, one has to be careful, since you could have the ambiguous case.
One of the angles could be obtuse.
The arcsine will of course give you the corresponding acute angle.
e.g. sin 120 = .5 and sin 30 = .5
when you do arcsine(.5) on your calculator it will only give you 30º
A good rule of thumb is to use the Cosine Law to find the largest angle of the triangle, and avoid the problem, since there can be only one obtuse angle in the triangle
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