"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.

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- Mar 24th 2011, 06:36 AM[nuthing]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. - Mar 24th 2011, 06:46 AMPlato
- Mar 24th 2011, 01:40 PMZap
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.