Need help for questions in chapter one

[Hielly]

The formula states: Tan (theta) = M(gradient) The angle is the angle the line makes with the POSITIVE side of the x axis.

Using basic triangular angle calculation, you should get tan(105) = M (now this should be a negative value, as tan is negative in 2nd quadrant).

yups that correct!

so how do you get to 105 ?

edit: 90--15 ?

[mullums1]

hrm strange Tan(-15) = 0.856,
but the back of the book says -3.73

Maybe your calculator is in the wrong mode.

okay its in degrees but now it says
tan(-15)=-0.27

I think thats the wrong angle. Just read my previous post, you should be getting 105.

Edit: Kool!!!

[Hielly]

mullums how did you get to 105 ?

[Flaming_Arrow]

sorry i forgot how the formula states it should be with the positive x axis

[mullums1]

Look at the attachment, thats the angle you are looking for.......

Edit: i know the graph doesnt cross at (0,0) sorry  , but using just straight forward triangular angle calculations, you should get 105. As 15 is one of the angles, and the other is a right angle. so the angle in traingle would be 75. But we are looking for the angle on the opposite side of that (positive side of x axis) so that u sud subtract this from 180 to give you 105.

[Hielly]

okay got it

[mullums1]

7)Find the equation of the line which passes through the point of intersection of the lines whose equations are 7x – 3y – 19 = 0 and 3x + 2y + 5 = 0, given that the required line is parallel to the line with equation –5x – 2y = 3.


First, try to find out the point of intersection using the first 2 equations and working on them simulataneously.

You should figure out, that they meet at x=1 using the valus.

Now just sub x=1 into one of the equations to find the y value where they intersect.

Now you got this mysterious graph parallel to the 3rd equation. This means they share the same gradient.

Just make y the subject for the 3rd equation and find gradient (pronumeral in front of x).

So you got the x, y and m value of this graph.

Sub it into linear formula, y= mx + c , to find c and consequently the whole equation.

[Hielly]

ye thanks mullums, i was lost at the bit, where you had to sub x=1 into parralel equation but its all good now, so thanks!!

[Hielly]

anyone tend to give me a hint on this one?
2)     
For safety considerations, wheelchair ramps are constructed under regulated specifications. One regulation requires that the maximum gradient of a ramp exceeding 1200 mm in length is 1/14.
Does a ramp with gradient1/18 meet the specifications?

thanks

[polky]

anyone tend to give me a hint on this one?
2)    
For safety considerations, wheelchair ramps are constructed under regulated specifications. One regulation requires that the maximum gradient of a ramp exceeding 1200 mm in length is 1/14.
Does a ramp with gradient1/18 meet the specifications?

thanks

I'm not too sure about the wording of this one, but it seems as though it's asking if 1/18 is larger or smaller than 1/14.  Since 1/14 is the maximum gradient, anything with a gradient larger than 1/14 would not be approved!

[Hielly]

hrm im really confused of this question as well. however i checked the answer at the back and it said 'yes'

[Flaming_Arrow]

i think it means that a ramp can have a maximum gradient of 1/14, 1/18 is smaller than 1/14 so its approved.