Triangle leg-lengths and tangent of angles

M

Mathisconfusing101

Junior Member
Joined
Jun 19, 2006
Messages
52
sub20-15.jpg


Triangle ABC- 1. Length of leg opposite 63° angle
2. Length of leg adjacent to 63° angle
3. tan 63°

Triangle DFE 1. Length of leg opposite 63° angle
2. Length of leg adjacent to 63° angle
3. tan 63°

Triangle GJH 1. Length of leg opposite 63° angle
2. Length of leg adjacent to 63° angle
3. tan 63°

I know its long.. and I know I am asking for a lot of help on here, but I really appreciate it. I can help out my English and History and Child Development.. I just can't seem to get math..
 
I got it! yay!

Just to make sure I did it right...

Triangle ABC- 1. Length of leg opposite 63° angle 10
2. Length of leg adjacent to 63° angle 5.1
3. tan 63° 1.96

Triangle DFE 1. Length of leg opposite 63° angle 12.4
2. Length of leg adjacent to 63° angle 6.3
3. tan 63° 1.96

Triangle GJH 1. Length of leg opposite 63° angle 15.1
2. Length of leg adjacent to 63° angle 7.7
3. tan 63° 1.96
 
OK...you divided 10 / 5.1 and 12.4 / 6.3 and 15.1 / 7.7 to get 1.96 in each case:
BUT what did you learn from that?

Take the smaller triangle ABC:
what is the length of AC?
what is size of angle at C?

AND: how are your 3 triangles related, since they all have same angle size?
 
I think the length of AC would be the A^2 + b^2 = C^2 thing would come up rite? and the angle would be 27* (degrees lol) rite?
 
Yes you would want to use Pythagora's Thereom in which a&b are the legs of the right triangle and c is the hypotenuse (the longest side in a right triangle directly across from the 90 degree angle). Can you complete Dennis's question w/this new knowledge?
Dennis said:
what is the length of AC?
Click to expand...
 
To calculate AC yes you would use Pythagorean Thereom:

\(\displaystyle \L (a)^2+(b)^2=(c)^2\) In which \(\displaystyle \L\bold a\) and \(\displaystyle \L\bold b\) are the legs on the right triangle and \(\displaystyle \L\bold c\) is the hypotenuse.

\(\displaystyle \L (5.1)^2+(10)^2=(c)^2\)

\(\displaystyle \L (26.01)+(100)=(c)^2\)

Can you finish?
 
You must log in or register to reply here.
Top