length of x

Printable View

• April 2nd 2008, 07:05 AM
Jamie88
length of x
http://users.cs.cf.ac.uk/J.G.Lee/project_html/shape

How can I find x?

I have examples but they are all triangles. This isn't a triangle so I don't know. :s
This is so easy but my notes don't cover it. :|

I have tried 15 / 1.8 but this surely can not be right?

Thanks in advance. :)
• April 2nd 2008, 08:21 AM
nervousnelly
create a triangle by drawing a horizontal line at the bottom of the 1.2m side on the left. Then you know that line is 15m and you can work out the vertical side on the right. Then you've got a right angle triangle where you know two sides, and I guess you work out angles and whatever.
• April 2nd 2008, 08:31 AM
Jhevon
nervousnelly is right. you would of course use similar triangles to solve for x.

(base of large triangle)/(height of large triangle) = (x)/(height of small triangle)
• April 2nd 2008, 09:05 AM
Jamie88
I have the base of the large triangle as 2.5m which I think is correct.
I am confused on how I'm meant to find x as it is a smaller triangle inside the large one.
• April 2nd 2008, 04:42 PM
Len
Quote:

Originally Posted by Jamie88
I have the base of the large triangle as 2.5m which I think is correct.
I am confused on how I'm meant to find x as it is a smaller triangle inside the large one.

It is indeed a smaller triangle but it does share angles so let A= the bottom angle

$Tan(A)=\frac{opp}{adj}=\frac{15}{3.7-1.2}=\frac{x}{1.8}$

$\frac{15}{3.7-1.2}=\frac{x}{1.8}$

Then just solve for x

$x=\frac{15*1.8}{2.5}=10.8$
• April 3rd 2008, 03:22 AM
Jamie88
Ohhh I see this now, thanks matey. :)