MATH.. triangles? help..?

Question

ok i have to find the lengths of the sides of 2 triangles..

i have a 30-60-90degree triangle and a 45-45-90degree triangle.

it says the hypotenuse of the triangles is one unit.. how do i find the lengths of the sides? it says no decimal approxamations, leave them in exact form.

Answer 1

The 45-45-90 right triangle,
You know that the 2 legs are equal to each other, so call them x.
Using the Pythagorean theorem:
x² + x² = 1²
2x² = 1
x = 1/(√2) = (√2)/2

The 30-60-90 right triangle:
Draw an equilateral right triangle with all sides being of length 1. Every angle is 60. Draw a perpendicular bisector through one of the angles. Now you have 2, 30-60-90 right triangles, but the base(the side opposite the 30˚ angle) of each 30-60-90 right triangle is of length 1/2. Now you have the hypotenuse of length one 1 and the leg opposite the 30˚ angle equalling 1/2. Now all you have to do is find the length of the side opposite the 60˚
Using the pythagorean theorem
1² = (1/2)² + x²
1 - (1/4) = x²
(3/4) = x²
√(3/4) = x
√(3)/2 = x

Answer 2

For the 30-60-90 triangle if the hy potenuse is 1 then the short leg is 1/2 and the long leg is sqrt(3)/2.

For the 45-45-90 triangle it the hypotenuse is 1 then each leg is = to Sqrt(2)/2.

Answer 3

Here is something you must remember and never forget, as you'll see it again and again, especially in trig class.

30-60-90 triangles have sides 1/2, sqrt(3)/2, and 1
45-45-90 triangles have sides sqrt(2)/2, sqrt(2)/2, and 1

Square them and you'll see the squares of the sides add up to the square of the hypotenuse.

Now, inscribe it in your brain like you memorized the multiplication tables. Trust me, you'll be encountering it a lot.

Answer 4

for 30-60-90 tri

sin 30 = 1/2 =Perp/hypo
1/2 = perp / 1
perp = 1/2

base = sqrt (1-1/4) ----> pythagoras theorem
base = sqrt (3/4) = sqrt3/2

Similarly do for 45-90-45