Trig
posted by Trig on .
h t t p : / / m a t h c e n t r a l . u r e g i n a . c a / Q Q / d a t a b a s e / Q Q . 0 9 . 9 9 / a n g e l a 2 . 2 . g i f
give that picture how do I solve for
(h) I know that you you
From the diagram tan(3.5) = h/(13 + x) and tan(9) = h/x. Solve the second equation for x, substitute in the first equation, and solve for h.
i\'m just having a hard time getting a formula for h if you could show me how to do it that would be great...
thanks...

http://mathcentral.uregina.ca/QQ/database/QQ.09.99/angela2.2.gif
ok,first look up tan(3.5)=.061163
so h= (13+x)*tan(3.5)
Then, tan 9 you can look that up
tan9=h/x
or h=x*tan9
setting h=h
(13+x)tan3.5=xtan9
solve for x
Finally, put that into the second equation you typed, and solve for h
h=x*tan9 
I would solve each equation for h, and then equate the two right sides ...
form #1
h = (13+x)tan3.5
from #2
h = xtan9
then :
(13+x)tan3.5 = xtan9
13tan3.5 + xtan3.5 = xtan9
13tan3.5 = xtan9  xtan3.5
13tan3.5 = x(tan9  tan3.5)
x = 13tan3.5/(tan9  tan3.5)
evaluate x and sub back into #2
or ...
in the left triangle we can find the other two angles, the one exterior to 9 would be 171, making the top angle 5.5ยบ
Now by the Sine Law you could find the side which is the hypotenuse to the rightangled triangle (call it b)
b/sin3.5 = 13/sin5.5
once you have that hypotenuse the other two sides of the rightangled triangle are easy to find
(Go UofR)