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# Triangle side lengths/trignometry

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For this problem, I found the height of the triangle first, which is $$\sqrt{8^2-5^2}=\sqrt{39}$$. I'm trying to find the length now and I just want some confirmation that what I'm doing is correct. I split triangle SVT with a line to divide it into two right triangles. Let the point where the the altitude intersects with the ST be X. The height of this triangle is $$\sqrt{39}$$. From S to X, using the pythagorean thereom, SX=19. Then from X to T, we know XT is 5. So the length of ST is 24. Is this correct?

Feb 9, 2021

#1
+116125
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Note that TU  = SW

So

WV  =  sqrt (SV^2  - SW^2)    =    sqrt (20^2    -  39)

So....you should be able to  finish it  now

Feb 9, 2021