question with the cosine rule, dont get it (1 Viewer)

[Fruitcake05]

Its question 3b) form the 2001 HSC.

Triangle ABC. AC = x, CB = 13, and CA = 7. Angle CBA = 60.

Use the cosine rule to show that x²-7x=120, and hence find escat value
of x.


13² = 7² + x² - 2.7.x.cos60degrees

i got to here, and not sure whats next

 

[Slidey]

Solve for x:

169-49=x^2-7x
x^2-7x-120=0
Inspection: 15, 8
(x-15)(x+8)=0
x=15 or -8
Discard x=-8, since negative distances don't exist

.'. Side x = AC = 15

 

[Fruitcake05]

Thanks slide, seems like your the maths legend on the boards.

Anyway I understand it now, but I just dont get what happens to the cos60degrees?

 

[Jago]

cos 60 = 1/2

when you multiply that by the 2, it disappears.

 

[Fruitcake05]

Ah yes of course............

And just one little question while I am here - I dont really pay attention during year 9 & 10, and I need to when I have to use inverse cos and when to use normal cos............. and what do i put into my calculator to get cos60deg= 1/2 ? I now know its half because i did the 30 and 60 deg triangle where you can see it.

 

[Slidey]

Thanks slide, seems like your the maths legend on the boards.

Well, a lot of other people help out around here, so I wouldn't say that. I do like maths, though.

 

[Slidey]

Ah yes of course............

And just one little question while I am here - I dont really pay attention during year 9 & 10, and I need to when I have to use inverse cos and when to use normal cos............. and what do i put into my calculator to get cos60deg= 1/2 ? I now know its half because i did the 30 and 60 deg triangle where you can see it.

Inverse cos is called 'arccos'.

Just remember that cos(60)=1/2

Thefore it must be the case that arccos(1/2)=60.

To get cos60=1/2, you punch in: cos(60), ENTER

To solve cos(x)=1/2, for example, you would enter:
arccos(1/2) and the answer would be x.

Arccos is denoted by cos^-1 on your calculator.

 

[Fruitcake05]

ok thanks guys. Gotta tackle the work now.