Finding Angle C in an ABC Triangle

[Natasha1]

Homework Statement

ABC is a triangle
The angle A = 30 degrees
AB = 7cm
BC = 5 cm
C is an acute angle. Find the size of angle C?

Homework Equations

I need to use the cosine rule here Cos C = (a^2 + b^2 - c^2) / 2ac

The Attempt at a Solution

The problem is I do not know the size of side b?
So how can I attempt to resolve the answer?

Cos C = (5^2 + b^2 - 7^2) / 2*5*7

 

[Ray Vickson]

Homework Statement

ABC is a triangle
The angle A = 30 degrees
AB = 7cm
BC = 5 cm
C is an acute angle. Find the size of angle C?

Homework Equations

I need to use the cosine rule here Cos C = (a^2 + b^2 - c^2) / 2ac

The Attempt at a Solution

The problem is I do not know the size of side b?
So how can I attempt to resolve the answer?

Cos C = (5^2 + b^2 - 7^2) / 2*5*7

Why do you need to use the cosine rule? Did somebody tell you that you are being forced to use it?

Just use whatever tool is the most convenient. If I were doing it I would let the points be at $A = (0,0), B = (7,0), C = (x,y).$ You can use the information about angle A and distance BC to get two equations in $x, y$. These can be reduced to a single equation in $x$, which is solvable without too much trouble.

 

[Natasha1]

Thanks Ray but we have never used this method in class and I am sure this is the approach my teacher wants us to do it. Can I not use the cosine rule or sine rule then?

 

[gneill]

One way or another you'll have to find side b if you're going to apply the cosine rule. If you're not comfortable jumping right into the algebra with a method you've not used before, then start with a diagram and come up with a plan from there.

 

[BvU]

I need to use the cosine rule here

Says who ?

 

[BvU]

Can I not use the cosine rule or sine rule then?

What about this second option ?

 

[Natasha1]

Ah ha!

cos A / a = sin C / c

So cos 30 / 5 = sin C / 7

So sin C = 7*(cos 30 / 5) = 1.212435565

How do I get the angle?

 

[BvU]

cos A / a = sin C / c

Not how I remember the law of sines

How do I get the angle?

The button on your calculator don't work for cosines > 1 ?

 

[Natasha1]

Ah ha!

sin A / a = sin C / c

So sin 30 / 5 = sin C / 7

So sin C = 7*(sin 30) / 5 = 0.7

arcsin gives 44.4 degrees

 

[neilparker62]

In general you will use the cos rule for solving a triangle if the givens are the 3 sides or alternatively 2 sides and an angle formed between the 2 given sides.

 

[Mark44]

Ah ha!

sin A / a = sin C / c

So sin 30 / 5 = sin C / 7

So sin C = 7*(sin 30) / 5 = 0.7

arcsin gives 44.4 degrees

It could be that there are two different triangles that satisfy the given measurements, as in the drawing in post #4. If $\sin(C) = .7$ there is also another angle that is possible for C.

Edit: Since it is stated in post #1 that C is an acute angle, the above does not apply.

 

[BvU]

C is an acute angle

 

[gneill]

It could be that there are two different triangles that satisfy the given measurements, as in the drawing in post #4. If sin(C)=.7sin⁡(C)=.7\sin(C) = .7 there is also another angle that is possible for C.

Actually, that was my primary reason for posting the diagram. Too subtle?

 

[Mark44]

C is an acute angle

Missed that in the first post.

 

[gneill]

Maybe too subtle, since the OP didn't seem to pick up on it.

Okay, will a tad more blatant in future. I hate blatant when gently nudging

 

[SammyS]

Actually, that was my primary reason for posting the diagram. Too subtle?

I thought that it was just subtle enough.

 

[symbolipoint]

I drew the diagram according to original problem description, and intuition tells me that b, and angle B, and angle C are not locked into any set of single values. Not a unique triangle. That's only first impression.

Thinking further,
forget about trying Law Of Cosines.

Law Of Sines seems fitting:

sin(C)/7=sin(30)/5

sin(C)=(7/5)sin(30)

sin(C)=7/10
and from this you could get value for angle C.Now, you could use AB, BC, and angle at B, to find b, using Law Of Cosines.; but you could use Law Of Sines again if you want.

 

[BvU]

I drew the diagram according to original problem description, ...

Why come with this now, when OP has cracked the case in #9 ?

 

[symbolipoint]

Why come with this now, when OP has cracked the case in #9 ?

Post was not old. I had to try the problem myself to see if or where Law Of Cosines would be useful; and I edited my own post twice within 20 minutes. Post #9 did not indicate any use of Law Of Cosines.

 

[BvU]

Thanks Ray but we have never used this method in class and I am sure this is the approach my teacher wants us to do it. Can I not use the cosine rule or sine rule then?