1. Geometry Honors

Hello all;

I need some help with the following:

1) Line L intersects the x axis going towards the top right hand corner of the grid. It forms a 60 degree angle to the right of the point of intersection. What is the slope of the line?

2) A trapezoid has the following measurements:
Top base: 30
Left side 15
Right side 24

A line is drawn form each of the top angles, and meet the base perpendicularly. (You now have 2 right triangles) The triangle on the right has a 60 degree angle on the bottom right of it. What is the measure of the ENTIRE bottom base?

We are doing 60-30 triangles, Pythagorean thrum and triples, etc, so this will most likely be used!
'Thankyou!

2. 1) Line L intersects the x axis going towards the top right hand corner of the grid. It forms a 60 degree angle to the right of the point of intersection. What is the slope of the line?

If you know the angle of a line, one way to figure out the slope is to use results from trigonometry. You can draw a right triangle whose hypotenuse lies on the line: pick a point on the line and draw a vertical line to the x-axis then take the point where the line intersects the x-axis and draw a horizontal line along the axis over to the point where the vertical line hits the x-axis.

We can say that the hypotenuse of this triangle has a length of h (its pretty cool that it won't actually matter how long it is). Then the base has a length
h*cos(60), and the height has a length of h*sin(60) and the slope is the height divided by the length:

slope = h*sin(60)/h*cos(60) = sin(60)/cos(60) = sqrt(3) (see how the h drops out).

If you haven't learned about sin and cos functions but you know properties of a 60-30 triangle you should know that a 30-60 triangle has side length ratios 1,2,sqrt(3). So the slope of the line is again the length of the opposite side (opposite 60) divided by the adjacent side (rise over run) which is sqrt(3)/1 = sqrt(3).

For problem 2)
remember that the side opposite the angle-60 has a length twice the side adjacent to the angle-60, and the hypotenuse has a length sqrt(3) times the length of the side adjacent to the angle-60.

3. iknowone, for Q1: can't u just do Tan(60) ?

Q2: If this is what were dealing with? im a bit confused..

to find y we 15^2 - x^2. but i work x out to be 20.78 by 24*Sin(60)...

meaning y is complex!?

4. Originally Posted by Stone Cold
iknowone, for the first, can't u just do Tan(60) ?
Yes, but how do you know you can do that? I wouldn't assume anyone knows that formula. I didn't until I was in college.

-Dan

5. Originally Posted by topsquark
Yes, but how do you know you can do that? I wouldn't assume anyone knows that formula. I didn't until I was in college.

-Dan
I haven't got around to changing my country flag, i actually live in Spain and do IGCSE (UK thing) so Ive learnt different things at different times.

6. Nice picture Stone Cold. First solve for $x$ using the fact that $z = \frac{24}{2} = 12$ and the opposite side is $\sqrt{3}$ times the base length in a 30-60-90 triangle.

Then you can use the Pythagorean theorem to solve for $y$ in the second right triangle.

7. Iknowone,

to find x i can do, 24*Sin(60) or (and thanx cos this is the first i heard of this) $(1/2*24)*sqr(3)$

Now, armed with x, which is 20.78...

"y" is then $15^2 - 20.78^2 = -206$...

meaning to get the $sqr(y^2)$ it will be complex!?

8. Ah ha so perhaps the drawing was not so accurate since 15 is the legnth of the hypotenuse it must be that x < 15. Perhaps it was a typo and the triangle with hypotenuse length 15 has the 60 degree angle?

9. RIGHT!, In this case, the base of the triangle is equal too... 20.2

so that the base of the trapesium is...

20.2+30+7.5 = 57.7