Posted by Trig on Thursday, August 27, 2009 at 7:36pm.
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 5.5º
Now by the Sine Law you could find the side which is the hypotenuse to the right-angled triangle (call it b)
b/sin3.5 = 13/sin5.5
once you have that hypotenuse the other two sides of the right-angled triangle are easy to find
(Go UofR)
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