"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.
"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.
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.