Thursday

May 5, 2016
Posted by **Mike** on Saturday, October 28, 2006 at 8:16am.

I have subsequently found a converter on the net which takes all the headache out of arriving at the correct answers.

The thing is I still do not understand how to solve a right triangle without the converter and need more help.

I have found other instructions that tell me to use a scientific calculator but I would like to be able to do it with a pen & paper.

Perhaps the calculations are horrendous.

Please help.

Thank you,

Mike.

What to you mean by "solve" a right triangle? The Pythagoran theorem will tell you one side length if you know the other two. If you know one acute angle besides the right angle, then the third angle is 90 degrees mnus the other acute angle. If you know one side length, you can get others using sine and cosine functions.

Thanks.

I know 1 side and 2 angles.

Say the side is 600mm, one angle is 90 degrees and another angle is 53 degrees. Clearly the other angle will be 37 degrees but how do I establish the lengths of the other sides without resorting to computer or calculator.

An example posted based on the above would be most helpful.

Thank you

Mike.

With two angles known, you really know all three. You need to use a sine or cosine relationship to get the other sides. This would be true using eight the law of sines

sin A/a = sin B/b = sin C/c = 1/c,

or just the definition

a/c = sin A

In either case, you need a calculator or a table to get the sine function, or you could use an infinite series to calculate the sin.

Thank you drwls.

Have printed a table of sine off the net and will now try to fathom out how to use it.

Mike

I know 1 side and 2 angles.

Say the side is 600mm, one angle is 90 degrees and another angle is 53 degrees. Clearly the other angle will be 37 degrees but how do I establish the lengths of the other sides without resorting to computer or calculator.

An example posted based on the above would be most helpful.

Even if using a calculator, you cannot find the other sides since you do not specify which angle the 600mm is opposite.

You might be surprised to discover that the two other sides of a Pythagorean triangle can be found knowing only one side and no angles.

The most general formulas for deriving all integer sided right-angled Pythagorean triangles, have been known since the days of Diophantus and the early Greeks. For a right triangle with sides X, Y, and Z, Z being the hypotenuse, the lengths of the three sides of the triangle can be derived as follows: X = k(m^2 - n^2), Y = k(2mn), and Z = k(m^2 + n^2) where k = 1 for primitive triangles (X, Y, and Z having no common factor), m and n are arbitrarily selected integers, one odd, one even, usually called generating numbers, with m greater than n. The symbol ^ means "raised to the power of" such that m^2 means m squared, etc.

Given any positive integer, it is possible to derive a Pythagorean triple with that integer as one of the sides.

For odd integers, given the integer X; resolve X into any two factors; equate the larger factor to m + n and the smaller to m - n; solve for m and n; define the triple using X = k(m^2 - n^2), Y = k(2mn), and Z = k(m^2 + n^2) where k = 1 for primitive triangles

For even integers, given the integer X; equate X to 2mn; let m and n can be any two numbers that yield the product 2mn; derive the triple using X = k(m^2 - n^2), Y = k(2mn), and Z = k(m^2 + n^2) where k = 1 for primitive triangles.

Od course, if you are given an angle, or angles, you must use the sin, cos or tan functions which are only available on a calculator.

Disregarding the angles for the moment, you can equate your side of 600mm to 2mn making mn = 300. Therefore, m and n can be any two numbers with a product of 300.

Just as an example, let m = 20 and n = 15. Then, assuming k = 1, Y = 2(20)15 = 600, X = 20^2 - 15^2 = 175 and Z = 20^2 + 15^2 = 625.

Therefore, 175^2 + 600^2 = 625^2, a righttriangle but not necessarily with the angles you specify. You could check the angles with a calculator.

There are many other Pythagorean triangles derivable with a side equal to 600mm, but which ones, if any, have the angles you specify, cannot be found without a calclator.

Whatever the integer angles, the sides cannot be integers also.

Have you heard of the indian named "SOH CAH TOA"?

Same her indian name out loud and memorize it.

sin = O / H

cos = A / H

tan = O / A

- Math -
**philen**, Sunday, November 27, 2011 at 12:42pmis this high geometry?

- Math -
**soh cah toa**, Sunday, November 27, 2011 at 12:43pmthats my name!