Posted by **Trig** on Thursday, August 27, 2009 at 7:36pm.

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...

- Trig -
**bobpursley**, Thursday, August 27, 2009 at 7:43pm
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

- Trig -
**Reiny**, Thursday, August 27, 2009 at 7:55pm
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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