# third side

posted by .

If one angle and two sides of a triangle are given, how to find the third side?

This is for solving a 9th grade geometry problem of finding the distance from earth to venus

You can use a standard formula for the side of the triangle that must be mentioned somewhere in your math book:

c^2 = a^2 + b^2 - 2 a b cos(alpha)

Here a and b are thelengths of the two sides of a triangle and alpha is the angle between these two sides.

Thanks. I know that formula, but I am trying to help out a ninth grader who says he doesn't know(haven't learned) that formula.
This problem comes under geometry, where you learn SSS, SAS, AAS postulates, and the question is about two different positions of Venus where distance between Sun and Venus(1.1 * 10^8km) & Sun and Earth(1.5 * 10^8km) are given. One angle(30degree) also is given. Need to find the distance between Venus and Earth.

Thanks again.

Hi Jen,

I see! The angle of 30 degrees, is this between the lines Sun-Earth and Sun-Venus, or one of the two other angles?

Hi Jen,

I see! The angle of 30 degrees, is this between the lines Sun-Earth and Sun-Venus, or one of the two other angles?

It is between sun-earth and earth-venus lines.

I see!

Well, to be honest, I've forgotten almost everything I learned about geometry in high school. This is how I would solve this problem.

Denote the position of the Earth relative to the Sun as E (E is a vector) and the position Venus Relative to the Earth as V. We know that the angle between the vectors E and V is (180 - 30) degrees is 150 degrees.

The position of Venus relative to the Sun is:

VS = E + V

Take the square of both sides (squaring a vector means taking the inner product with itself):

VS^2 = E^2 + V^2 + 2 E dot V

E dot V is the inner product of E and V, which is the length of E times the length of V times the cosine of the angle between E and V. So, you find:

|V|^2 -Sqrt(3)|E||V|+|E|^2 - |VS|^2 =0

Here |V| is the distace Earth-Veus we want to solve, |E| is the known distance Sun-Earth, and |VS| is the known distance Sun-Venus.

So, you have a quadratic equation for |V|. I'm not sure if there is a simple geometric reasoning appropriate for 9-th grade that will lead you to this equation. When I was in high school I was far ahead with my maths and knew about inner products etc. so I never bothered using complicated geometric reasoning using triangles if I didn't have to :)

This has to simple.
I wish I could post that picture so you will get a clear idea.

This has to simple.
I wish I could post that picture so you will get a clear idea.

If the angle is 30 degrees then there are two solutions:

2.1*10^8 km and 0.49*10^8 km. This is what you would expect, because the line 30 degrees from the line Earth Sun will intersect the orbit of Venus at two points.

How did you get that? Those are close(answers given are 2.0 *10^8km and 0.5*10^8km)
Thanks

There are two possible positions for Venus for a given position of Earth. This is an SSA combination with the sun, earth, venus as the vertices of the triangle.
The radius of earth's orbit is 1.5 * 10^8 kilometers, and the radius of venus' orbit is 1.1 * 10^8 kilimeters.

The angle marked 30 degree.

With all these information, how do I find out the distance from earth to venus for both positions of venus? (The answers are 2 * 10^8km, and 0.5 * 10^8km)

Is there a way I can send you an image?
Is there a way we can solve it using a congruence postulate?

Hi Jen,

I solved the quadratic equation I gave above. If you want to send me the picture you would have to know my email address, but I can't post that here.

I have a blog that starts with my name on blogspot. If you can find it and post a message there we can settle this.

You can type the two keywords:

iblis qualia

in google (without quotes) to find the blog.

The Law of cosines will provide you with what you want, the distance between earth and venus.

Given triangle VSE with angle VSE = 30º, (VE)^2 = (VS)^2 + (ES)^2 - 2(VS)(ES)cos(30).

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