# Trig Problem

1. Dec 28, 2004

### 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.

$$a^2=b^2+c^2-2bc cos(A)\longrightarrow 18.49=27.04+56.25-78 cos(A)$$

I tried bringing everything to the left side with the exception of cos(A) and then doing $$cos^-1$$ but it just wont work in my calculator I keep getting a math error.

Last edited: Dec 28, 2004
2. Dec 28, 2004

### 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.

Last edited: Dec 28, 2004
3. Dec 28, 2004

### Cantari

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

4. Dec 28, 2004

### VietDao29

Hi,
Dextercioby, haven't you realised that your post is completely wrong?
"Has it ever occured 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,

5. Dec 28, 2004

### The Bob

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 ©)

6. Dec 28, 2004

### BobG

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.

Last edited: Dec 28, 2004
7. Dec 28, 2004

### 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

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