Solving a Triangle with the Cosine Law: Help!

[aisha]

I just read about the cosine law and the sine law.

I have a practise problem and know to use the cosine law but what ever answer I get gives me a math error in my calculator.

the sides are a=4.3 b=5.2 c=7.5 I need to solve the triangle so find the 3 angles within.

a^2=b^2+c^2-2bc cos(A)\longrightarrow<br /> <br /> 18.49=27.04+56.25-78 cos(A)

I tried bringing everything to the left side with the exception of cos(A) and then doing cos^-1 but it just won't work in my calculator I keep getting a math error.

 

[phreak]

Are you sure you're on degree mode? Trig is usually only used with radians, but this one should involve degrees.

Isolate the cosA

-64.8 = -78(cosA)
Divide by -78.

cos^-1(.83) should yield the answer, but if you're on radians, it probably won't work.

I got 33.82 as an answer.

 

[Cantari]

Must just be an arithmetic error. I don't think it is a radian/degree issue.

 

[VietDao29]

Hi,
Dextercioby, haven't you realized that your post is completely wrong?
"Has it ever occurred to u that one side (viz."c") is exactly the sum of the other two...?I guess not,else u have realized that your triangle is not a regular one.It's a degenerate triangle.It has one angle of 180° and the other of 0°.The three summits are on the same line."
With a=4.3 b=5.2 c=7.5
Are you trying to say that c = a + b? And therefore 7.5 = 4.3 + 5.2 ?!
So... A = 33.82 is the answer.
Bye bye,
Viet Dao,

 

[The Bob]

I just read about the cosine law and the sine law.

I have a practise problem and know to use the cosine law but what ever answer I get gives me a math error in my calculator.

the sides are a=4.3 b=5.2 c=7.5 I need to solve the triangle so find the 3 angles within.

a^2=b^2+c^2-2bc cos(A)\longrightarrow<br /> <br /> 18.49=27.04+56.25-78 cos(A)

I tried bringing everything to the left side with the exception of cos(A) and then doing cos^-1 but it just won't work in my calculator I keep getting a math error.

Try rearranging the cosine equation:

a^2=b^2+c^2-2bccosA

=> a^2+2bccosA=b^2+c^2

=> 2bccosA=b^2+c^2-a^2

=> cosA=\frac{b^2+c^2-a^2}{2bc}

Now try adding the numbers to this:

cosA = \frac{(5.2^2)+(7.5^2)-(4.3^2)}{(2 \times 5.2 \times 7.5)}

cosA = \frac{27.04+56.25-18.49}{78}

cosA = \frac{64.8}{78}

cosA = 0.8307

A = cos^-^10.8307

A = 33.82

Just do that for the rest (but with a = 5.2, b = 7.5 and c = 4.3 etc...) and you will have three angles for the triangle.

Hope that helps.

The Bob (2004 ©)

 

[BobG]

I just read about the cosine law and the sine law.

I have a practise problem and know to use the cosine law but what ever answer I get gives me a math error in my calculator.

the sides are a=4.3 b=5.2 c=7.5 I need to solve the triangle so find the 3 angles within.

a^2=b^2+c^2-2bc cos(A)\longrightarrow<br /> <br /> 18.49=27.04+56.25-78 cos(A)

I tried bringing everything to the left side with the exception of cos(A) and then doing cos^-1 but it just won't work in my calculator I keep getting a math error.

Just out of curiosity, for cos^{-1}, are you hitting:

2ND key, COS key

or are you hitting:

COS key, 2ND key, x^{-1}

cos^{-1} is an abbreviated term for ARCCOS. Above your COS key, you should either have ARCCOS or cos^{-1}.

The only other possible problem is if you entered the equation into your calculator wrong. You can't take the arcosine of a number larger than 1 and that will also give you an error.

 

[aisha]

I got it I was subtracting 78 instead of dividing both sides and isolating cosA

My answers for the angles are <A=34 <B=42 and <C=104