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April 20, 2014

April 20, 2014

Posted by **Sha** on Saturday, March 14, 2009 at 11:48am.

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^2-2bc Cos A

16^2=13^2+10^2-2(13)(10) Cos A

16^2-13^2-10^2/-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

- Algebra II/ Trig -
**bobpursley**, Saturday, March 14, 2009 at 11:56amWhay is A 152deg? Why not 87 deg?

- Algebra II/ Trig -
**Sha**, Saturday, March 14, 2009 at 11:57ami don't know

can you explain step by step

- Algebra II/ Trig -
**bobpursley**, Saturday, March 14, 2009 at 12:07pmHuh?

cosA=.05

A= arccos.05

Use your calculator. Think. When is cosine of an angle near zero?

- Algebra II/ Trig -
**GanonTEK**, Saturday, March 14, 2009 at 12:07pmI'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.

- Algebra II/ Trig -
**bobpursley**, Saturday, March 14, 2009 at 12:09pmGanon, giving answers seldom helps a student, at least in my experience.

- Algebra II/ Trig -
**Sha**, Saturday, March 14, 2009 at 12:16pmThe procedure that I did is correct just the math is incorrect...is this what you are telling me?

- Algebra II/ Trig -
**Reiny**, Saturday, March 14, 2009 at 2:11pmWhen 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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