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Solve the following values

In triangle ABC:
Angle C is 120 degrees, angle A is 30 degrees, angle B is x degrees. Side c is 20cm, side b is ycm, side a is zcm.

Angle X= 180-(30+120) = 30 degrees.
As angle X=angle A the triangle is isoleces.

I split the triangle through the line of symetry through angle C so to solve y:
I now have angle C=120/2=60 degrees, side c is 10cm. Use trig:

CosC=b/c
Therefore Cos60=y/10

So y=5

I can't seem to get the correct answer which is 11.5 (3 s.f).
What am I doing wrong?

Cheers

Reply 1

mazboy008

The triangle is not an equilateral triangle since no side is 90. It is a isosceles :smile:

.

To work out y and z (since y =z):
y = 10/cos30 = 11.5 (3sf)
Hope this helps

Reply 2

Mr M

Study Forum Helper

You have labelled the sides incorrectly. You should be using sin 60 for a start.

Reply 3

dthomas86

Thanks for the fast response. What sides have a labelled wrong?
Why should I be using sin instead of cos? Actually shouldn't I be using tan because side c (10cm) is opposite angle C, and side y is adjacent to angle C.

Cheers.

Reply 4

TenOfThem

Original post by dthomas86

Have you drawn a diagram

Which side is your hypotenuse

Is the 10 opposite or adjacent to your 60 degrees

Reply 5

dthomas86

Yeah I have drawn a diagram the hypotenuse is the longest side so the longest side should be side c which is 10cm (if the the triangle is split in half), side y looks like it is adjacent to angle C (60 degrees) in the diagram. But also side c looks like it is opposite side c.
Seems really difficult to distingush each side.