# Maths-Cosine rule

#1
How do you do the cosine rule?
I know the equation, but please can someone give me an example of a question and show me the method?
Thank you
0
7 years ago
#2

Hope this helps
0
#3
(Original post by LeFeuilly)

Hope this helps
Hope what helps?
its just a square box?
0
7 years ago
#4
(Original post by RBalboa)
Hope what helps?
its just a square box?
Can you not see the image?
0
7 years ago
#5

Find the length of BC.

AnswerDid you get 4.86cm? If so, well done.
If not, remember to use the formula:
a2 = b2 + c2 - 2bc cos A
Substitute the values into the formula.
a2 = 72 + 32 - 2 × 3 × 7 cos 35
a2 = 58 - 42 cos 35
a2 = 23.5956
a = 4.86cm
0
#6
(Original post by LeFeuilly)
Can you not see the image?
No not really PM me
0
#7
(Original post by AyshaU)

Find the length of BC.

AnswerDid you get 4.86cm? If so, well done.
If not, remember to use the formula:
a2 = b2 + c2 - 2bc cos A
Substitute the values into the formula.
a2 = 72 + 32 - 2 × 3 × 7 cos 35
a2 = 58 - 42 cos 35
a2 = 23.5956
a = 4.86cm
It need to calculate the angle of ABC but all the question gives me is a length of 8.1cm 7.5cm and the angle ACB 30degrees
0
#8
http://imgur.com/xrLTp7J
use this, it's my question
0
7 years ago
#9
(Original post by RBalboa)
No not really PM me
Ehhh I'll just type it up:

82°
/\
5cm (c)/ \ 6cm (b)
/____ \
a cm

Find "a" correct to 3 s.f.

a²=b²+c²-2bc cosA
a=√6²+5² -2 (6)(5) cos82
a= 7.26 cm
0
7 years ago
#10
(Original post by RBalboa)
How do you do the cosine rule?
I know the equation, but please can someone give me an example of a question and show me the method?
Thank you
It's basically just Pythagoras. If you take a non-rightangled triangle and drop a perpendicular from one vertex to get two right-angled triangles, and apply Pythagoras to one of them and use the definitions of cos and sin on the other one, you get the cosine rule. There's no magic about it; you just use it in the same kinds of ways you'd use Pythagoras.
0
#11
(Original post by LeFeuilly)
Ehhh I'll just type it up:

82°
/\
5cm (c)/ \ 6cm (b)
/____ \
a cm

Find "a" correct to 3 s.f.

a²=b²+c²-2bc cosA
a=√6²+5² -2 (6)(5) cos82
a= 7.26 cm
Look above at the link I sent that's what my question looks like
0
#12
(Original post by Smaug123)
It's basically just Pythagoras. If you take a non-rightangled triangle and drop a perpendicular from one vertex to get two right-angled triangles, and apply Pythagoras to one of them and use the definitions of cos and sin on the other one, you get the cosine rule. There's no magic about it; you just use it in the same kinds of ways you'd use Pythagoras.
err okay I think
0
7 years ago
#13
(Original post by LeFeuilly)
Ehhh I'll just type it up:

82°
/\
5cm (c)/ \ 6cm (b)
/____ \
a cm

Find "a" correct to 3 s.f.

a²=b²+c²-2bc cosA
a=√6²+5² -2 (6)(5) cos82
a= 7.26 cm
If you're using EDEXCEL iGCSE Mathematics A book 2, it's Exercise 87 question one (and sorry about the whole image fiasco - I can't send anything by mail -_-")
0
#14
(Original post by LeFeuilly)
If you're using EDEXCEL iGCSE Mathematics A book 2, it's Exercise 87 question one (and sorry about the whole image fiasco - I can't send anything by mail -_-")
I'm not but if u click the link I sent you then my question will come up
0
7 years ago
#15
(Original post by RBalboa)
I'm not but if u click the link I sent you then my question will come up
So, like the people above me have said, cosine rule is basically just substitution + pythagoras. In regards to your question, you need to use the SINE rule to answer part a. As for part b, you need 1/2 ab sin c to solve that.

Hope that helps
0
7 years ago
#16
(Original post by RBalboa)
How do you do the cosine rule?
I know the equation, but please can someone give me an example of a question and show me the method?
Thank you
You can use the cosine rule predominantly when you're given two sides and an included angle (an included angle means an angle which is subtended between the two lengths.)
0
#17
(Original post by iMacJack)
You can use the cosine rule predominantly when you're given two sides and an included angle (an included angle means an angle which is subtended between the two lengths.)
I cant use the sine rule because i have a length and angle and another lenghh but the angle isnt included between both lengths
0
7 years ago
#18
(Original post by RBalboa)
I cant use the sine rule because i have a length and angle and another lenghh but the angle isnt included between both lengths
I give up trying to explain this well with just text.

So:

If that fails to work: http://tinypic.com/r/15rxede/8
0
7 years ago
#19
(Original post by RBalboa)
It need to calculate the angle of ABC but all the question gives me is a length of 8.1cm 7.5cm and the angle ACB 30degrees

I don't understand what you are trying to say. You wanted a question and I gave you one with the method what are you still confused on
0
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