How do you find the side length of a triangle? - Answers

The most popular triangle used in construction, engineering and mathematics is the right triangle, which has a 90o angle at the base. We'll call your base "x", height "y" and diagonal (hypotenuse) "r".You can find these values using your calculator's trigonometric functions, or you can use the Pythagorean Theorem.Finding Values (Pythagorean Theorem):The Pythagorean Theorem is straightforward and states that, for a right triangle, r^2 = x^2 + y^2. We can use a little bit of algebra to find x, y and r.r = sqrt(x^2 + y^2)r^2 = x^2 + y^2r = sqrt(x^2 + y^2); remove the ^2 from rx = sqrt(r^2 - y^2)r^2 = x^2 + y^2r^2 - y^2 = x^2; move y^2 to the left side of the equationsqrt(r^2 - y^2) = x; remove the ^2 from xy = sqrt(r^2 - x^2)r^2 = x^2 + y^2r^2 - x^2 = y^2; move x^2 to the left side of the equationsqrt(r^2 - x^2) = y; remove the ^2 from yFinding Values (Trigonometric Functions):Acquiring Ratios:cos(angle) = x/rsin(angle) = y/rtan(angle) = y/xAcquiring Angles:cos-1(x/r) = anglesin-1(y/r) = angletan-1(y/x) = angleThat looks confusing. What are cos, sin, tan, etc?They're functions. You give a function input and it outputs something else. How they do this is complicated and required calculus.x = cos(angle) * rcos(angle) = x/rcos(angle) * r = x; move r to the left side of the equationy = sin(angle) * rsin(angle) = y/rsin(angle) * r = y; move r to the left side of the equationx = 1 / (tan(angle)/y)tan(angle) = y/xtan(angle) / y = 1/x; move y to the left side of the equation1 / (tan(angle) / y) = x; flip the equationy = tan(angle) * xtan(angle) = y/xtan(angle) * x = y; move x to the left side of the equationr = 1 / (cos(angle)/x)cos(angle) = x/rcos(angle) / x = 1/r; move x to the left side of the equation1 / (cos(angle)/x) = r; flip the equationr = 1 / (sin(angle)/y)sin(angle) = y/rsin(angle) / y = 1/r; move y to the left side of the equation1 / (sin(angle)/y) = r; flip the equation