Calculating an angle of a none right angle triangle

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

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

Re: Calculating an angle of a none right angle triangle

Quote:

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 $\displaystyle a=12,~b=22,~\&~c=28$. If that is so then

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

Re: Calculating an angle of a none right angle triangle

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)

Re: Calculating an angle of a none right angle triangle

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

Re: Calculating an angle of a none right angle triangle

Quote:

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.

Re: Calculating an angle of a none right angle triangle

hi plato

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

Re: Calculating an angle of a none right angle triangle

Quote:

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?

Re: Calculating an angle of a none right angle triangle

Quote:

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

$\displaystyle c^2=a^2+b^2-2ab\cos(\theta)$

He is clearly confused by the notation.

Re: Calculating an angle of a none right angle triangle

Quote:

Originally Posted by

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

If you say that $\displaystyle a = 12,~ b = 22 ~\&~ c = 28$

then $\displaystyle m(\angle B) = \arccos \left( {\frac{{12^2 + 28^2 - 22^2 }}{{2(12)(28)}}} \right)$

Re: Calculating an angle of a none right angle triangle

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

Re: Calculating an angle of a none right angle triangle

You say:

Quote:

Originally Posted by

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

But then you use

Quote:

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.

Re: Calculating an angle of a none right angle triangle

Quote:

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

Re: Calculating an angle of a none right angle triangle

Quote:

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.

$\displaystyle \angle B$ is opposite side $\displaystyle \overline{AC}$ so $\displaystyle b=28$.

$\displaystyle \angle A$ is opposite side $\displaystyle \overline{BC}$ so $\displaystyle a=22$.

$\displaystyle \angle C$ is opposite side $\displaystyle \overline{AB}$ so $\displaystyle c=12$.

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

Re: Calculating an angle of a none right angle triangle

Quote:

Originally Posted by

**Plato** I think that I now understand you confusion.

$\displaystyle \angle B$ is opposite side $\displaystyle \overline{AC}$ so $\displaystyle b=28$.

$\displaystyle \angle A$ is opposite side $\displaystyle \overline{BC}$ so $\displaystyle a=22$.

$\displaystyle \angle C$ is opposite side $\displaystyle \overline{AB}$ so $\displaystyle c=12$.

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

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

Re: Calculating an angle of a none right angle triangle

Quote:

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 $\displaystyle b^2=a^2+c^2-2ac\cos(B)$ for $\displaystyle \cos(B)$ we get $\displaystyle cos(B)={\frac{{a^2 + c^2 - b^2 }}{{2ac}}}$.