# Thread: lengths of sides in a right angled triangle

1. ## lengths of sides in a right angled triangle

"A right angled triangle with hypotenuse of length 6 contains the angle π/4. What are the exact lengths of the remaining two sides?"

my working leads me no where, any suggestions? thanks, any help will be appreciated.

2. Originally Posted by [nuthing]
"A right angled triangle with hypotenuse of length 6 contains the angle π/4. What are the exact lengths of the remaining two sides?"
If a right triangle contains an angle of $\frac{\pi}{4}$ then it is isosceles. So solve this equation: $2s^2=36~.$

3. I'm sorry, I'll will learn the latex editor as soon as I'm done here ... frown.. although Plato is is beyond brillient I wondered if I could offer a different and heavier handed approach. I wondered if your teacher wants you to resolve this with sin and cos functions.. if that is where you are in your book. If this is the case:

hypotenuse = 6
angle = pi/4 is a 45 degree angle
side adjacent is the leg on the x axis
side opposite is the leg perpendicular to the x axis

soooo, to get the side adjacent you use: cos (pi/4) = side adjacent / hypontenuse (this is the identity for a cosine)
rearrange here to get hypotenuse * cos (pi/4) = side adjacent
cos (pi/4) is sqrt(2)/2 and hyp = 6 So: 6*sqrt(2)/2=3*sqrt(2)

for the side opposite same deal, but you use the formula for sin: sin (pi/4)= side opposite / hyp (this is the identity for a sine)
giving: hyp * sin(pi/4)=side opposite
Now sin (pi/4) also equals sqrt(2)/2
so you will get the same answer: 6*sqrt(2)/2=3*sqrt(2)

they ask for exact lengths so you would leave the answers in the square root form.
in the end both legs then are 3*sqrt(2)

once again sorry for not using the editor.