Trig- Maths ext! (1 Viewer)

[marxisming]

In the triangle ABC, if a = 4cm, c= 2b-4 and cosC= 1/8

i. Use cosine rule to find b
ii. Find the perimeter of triangle ABC
iii. Find the area of the triangle ABC
iv. Prove that <C = 2 <A

If you could help me with this question, that would be great!
Thanks!!

 

[cwag]

1)..the trick in this question is to realise that the (a²) in the cosine rule is the oppostie of the angle. so after drawing the triangle, we can set up our equation as..

cos rule a²=b² + c² - 2bc cosA

(2b-4)²=b² + 4² - 8b*(1/8)......[note, cosC = 1/8, given in info]

4b² - 16b +16 = b² + 16
3b² - 15b = 0
3b(b - 5) = 0
b = 5 or 0
therefore b = 5

2) now we know b..the perimeter becomes 4 + 5 + (2*5 -4) = 15 units

3) Area = 1/2 absin C = 2*5*sin (cos^-1 1/8)
= aprox. 9.92 or for exact form.... (10*rt63)/8

4)im not quite sure how to prove it the way your spose to....but...
cos A = (6² + 5² - 4²) / (2*5*6)
cos A = 0.75
cos C = 0.125 (given)

therefore ratio A/C = cos^-1.75/cos^-1.125
=0.5
therefore <C = 2<A

 

[Sarah182]

First three are super easy, just use the cosine rule and from there you can work out the perimeter.
Use the area of a triangle rule for number 3.
Im not really sure about the last one, to be honest I dont understand the question which is a worry. *Leaves bos to revise trig*

 

[marxisming]

thanks all =)