Calculating an angle of a none right angle triangle

**David Green** · August 18th 2011, 02:39 PM

I am very new to maths so I would appreciate if posters are patient with me as this is a learning curve

I have a triangle with no equal sides and I am looking to find angle B. The length of the sides I will make up as 12cm, 22cm and 28cm.

I know all three sides and none of the angles and I am thinking that I should be using the cosine rule, therefore I have completed the following;

C2 = a2 + b2 – 2ab cos Ø

122 = 222 + 282 – 2 · 22 · 28 cos Ø
144 = 484 + 784 – 1232 cos Ø
144 = 1268 – 1232 cos Ø
1232 cos Ø = 1124
Cos Ø = 1124
1232
Cos Ø = 0.91
Cos -1 Ø = 24.2º (1dp)

Have I used the right rule, if not please advise why, and if I have would you please advise why I can't use the Sine Rule and what the difference is?

Thanks

David

**Plato** · August 18th 2011, 02:51 PM

Originally Posted by David Green

to find angle B. The length of the sides I will make up as 12cm, 22cm and 28cm.

First you must know which is which.
Say $a=12,~b=22,~\&~c=28$ . If that is so then
$m(\angle B) = \arccos \left( {\frac{{a^2 + c^2 - b^2 }}{{2ac}}} \right)$

**anonimnystefy** · August 18th 2011, 02:54 PM

hi David Green

you chose the right rule.the Sines rule wouldn't have worked because for that you need at least one angle.

i am not sure about the last line.it should be:
Ø=cos-1 (0.91)

**David Green** · August 18th 2011, 03:10 PM

This is confusing me now, if I use the first example I get the result 2.67, and if I use cos then I get 99.

If I use theta cos -1 (0.91) I get 24.5 degrees

Help

**Plato** · August 18th 2011, 03:16 PM

Originally Posted by David Green

This is confusing me now, if I use the first example I get the result 2.67, and if I use cos then I get 99.
If I use theta cos -1 (0.91) I get 24.5 degrees

Did you read reply #2?
The angle B is opposite the side with length b.
So it all depends on how you name the triangle.

**anonimnystefy** · August 18th 2011, 03:19 PM

hi plato

i think that's what confused the OP.you named it differently than him and me.

**David Green** · August 18th 2011, 03:20 PM

Originally Posted by Plato

Did you read reply #2?
The angle B is opposite the side with length b.
So it all depends on how you name the triangle.

Reply 2 is correct, A = 12, B = 22 and C = 28

So are we saying that the angle is 2.67 degrees?

2.67 = theta 99

0.91 cos -1 = 99

Are they not both the same then?

**Plato** · August 18th 2011, 03:24 PM

Originally Posted by anonimnystefy

i think that's what confused the OP.you named it differently than him and me.

No that is the whole point.
He did not name them at all.
He talked about finding B and the uses
$c^2=a^2+b^2-2ab\cos(\theta)$
He is clearly confused by the notation.

**Plato** · August 18th 2011, 04:01 PM

Originally Posted by David Green

Reply 2 is correct, a = 12, b = 22 and c = 28

If you say that $a = 12,~ b = 22 ~\&~ c = 28$
then $m(\angle B) = \arccos \left( {\frac{{12^2 + 28^2 - 22^2 }}{{2(12)(28)}}} \right)$

**David Green** · August 19th 2011, 09:53 AM

Looking deeper into this today it seems that I am using the wrong cosine rule?

The book in the summary section only shows; c^2 = a^2 + b^2 - 2ab cos C

I initially thought the book print had make an error where I thought that cos C should have read cos theta, which is what I then wrote on this forum.

Using;

b^2 = c^2 + a^2 - 2ac cos B

angle B =

28^2 = 12^2 + 22^2 - 2 x 22 x 12 cos B

784 = 144 + 484 - 328 cos B

- 528 cos B = 784 - 628

= 156 / 528

cos B = 0.29

theta = cos -1 (0.29) = 73 degrees.

This is what I think is right, any views welcome

David

**Plato** · August 19th 2011, 11:12 AM

You say:

Originally Posted by David Green

Reply 2 is correct, A = 12, B = 22 and C = 28

But then you use

Originally Posted by David Green

b^2 = c^2 + a^2 - 2ac cos B

28^2 = 12^2 + 22^2 - 2 x 22 x 12 cos B

Why change?

Do the calculations is reply #9 please.

**David Green** · August 19th 2011, 01:15 PM

Originally Posted by Plato

You say:

But then you use

Why change?

Do the calculations is reply #9 please.

The calculations in reply 9 would = 0.66 = 49 degrees

The problem there was that I used the wrong cosine rule.

if AB = 12 and BC = 22 and AC = 28, then I am asked to find the angle for B, then this is why I changed the cosine rule.

David

**Plato** · August 19th 2011, 01:24 PM

Originally Posted by David Green

The calculations in reply 9 would = 0.66 = 49 degrees

The problem there was that I used the wrong cosine rule.

if AB = 12 and BC = 22 and AC = 28, then I am asked to find the angle for B, then this is why I changed the cosine rule.

I think that I now understand you confusion.
$\angle B$ is opposite side $\overline{AC}$ so $b=28$ .
$\angle A$ is opposite side $\overline{BC}$ so $a=22$ .
$\angle C$ is opposite side $\overline{AB}$ so $c=12$ .

Now use $m\left( {\angle B} \right) = \arccos \left( {\frac{{a^2 + c^2 - b^2 }}{{2ac}}} \right)$

**David Green** · August 19th 2011, 01:30 PM

Originally Posted by Plato

I think that I now understand you confusion.
$\angle B$ is opposite side $\overline{AC}$ so $b=28$ .
$\angle A$ is opposite side $\overline{BC}$ so $a=22$ .
$\angle C$ is opposite side $\overline{AB}$ so $c=12$ .

Now use $m\left( {\angle B} \right) = \arccos \left( {\frac{{a^2 + c^2 - b^2 }}{{2ac}}} \right)$

Your right, the conclusion I get using the formula you wrote is 107 degrees?

Now I am even more confused

David

Edited section below;

I have had another look at the formula I used here;

b^2 = c^2 + a^2 - 2ac cos B

I can see in my transposition that I changed the negative sign to a positive when changing the integers from side to side because I thought ending up with a negative - 0.29 was incorrect, however after working out the angle I do indeed get 107 degrees, which is what I also get when using the formula as you wrote it, which I might add is a tidy method and proves that a less chance of making the mistake I made possible.

I think we can agree now that the solution is 107 degrees and I do like the way you do the formula, which is better than in my maths book.

Thanks

David

**Plato** · August 19th 2011, 01:42 PM

Originally Posted by David Green

Your right, the conclusion I get using the formula you wrote is 107 degrees? Now I am even more confused

The c107 is correct. What are you confused about?
If you solve $b^2=a^2+c^2-2ac\cos(B)$ for $\cos(B)$ we get $cos(B)={\frac{{a^2 + c^2 - b^2 }}{{2ac}}}$ .

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