Hi Jen,
I've seen the drawing. My site is down at the moment.
Draw a line from Venus (either of the two positions) to the line Earth-Sun at right angles.
Let's call the distance Earth Venus d. Then you have a right angle triangle with a hypotenuse d, the line that intersects the line Sun Earth at right angles has a length of
d sin(30°)= d/2 and the distance from the Earth to the point where that line of length d/2 intersects the line Earth Sun (let's call it P) is
d cos(30°) =1/2 sqrt(3)d
The distance from the sun to P is then 1.5*10^8 km minus 1/2 sqrt(3)d
Now the line Sun-Venus is the hypotenuse of another right angled triangle with sides Sun to P and the line of length d/2 you've drawn.
This means that
(1.5*10^8 km - 1/2 sqrt(3)d)^2 + (d/2)^2 = (distance from sun to venus) squared
You can then solve d from this quadratic equation.
To solve the quadratic equation and find the value of 'd', you will need to follow these steps:
Step 1: Rewrite the equation in a standard quadratic form, where you have one side equal to zero:
(1.5*10^8 km - 1/2 sqrt(3)d)^2 + (d/2)^2 - (distance from sun to venus)^2 = 0
Step 2: Expand and simplify the equation:
[(1.5*10^8 km)^2 - 2*(1.5*10^8 km)*(1/2 sqrt(3)d) + (1/2 sqrt(3)d)^2] + (d/2)^2 - (distance from sun to venus)^2 = 0
Step 3: Simplify further:
(2.25*10^16 km^2 - 1.5*10^8 km*sqrt(3)d + (3/4)d^2) + (1/4)d^2 - (distance from sun to venus)^2 = 0
Step 4: Combine like terms:
2.25*10^16 km^2 + (1/4)d^2 - 1.5*10^8 km*sqrt(3)d - (distance from sun to venus)^2 = 0
Step 5: Rearrange the equation in the form 'ax^2 + bx + c = 0':
(1/4)d^2 - 1.5*10^8 km*sqrt(3)d + (2.25*10^16 km^2 - (distance from sun to venus)^2) = 0
Now that we have the quadratic equation in standard form, where 'a' is 1/4, 'b' is -1.5*10^8 km*sqrt(3), and 'c' is (2.25*10^16 km^2 - (distance from sun to venus)^2), you can solve for 'd' using the quadratic formula:
d = [-b ± sqrt(b^2 - 4ac)] / (2a)
Plug in the values of 'a', 'b', and 'c' into the formula, and you can find the two possible solutions for 'd'. Remember to consider both the positive and negative square root when applying the ± sign.
Once you obtain the solutions for 'd', you can substitute them back into the equations you provided to determine the corresponding distances and complete the line drawing as instructed.