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August 2, 2015

Homework Help: trig

Posted by trig on Sunday, September 6, 2009 at 5:06pm.

h t t p : / / i m g 4 0 . i m a g e s h a c k . u s / c o n t e n t . p h p ? p a g e = d o n e& l = i m g 4 0 / 1 4 8 / a s d a s d y e . j p g & v i a = m u p l o a d

(def of tan theta = a^-1 o)a = o
opposite = adjacent tan theta

written with respect to the first angle

opposite = (adjacent 1 + adjacent 2) tan theta one

were adjacent 1 is the 50 feet

written for the second angle

opposite = adjacent 2 tan theta two

set them equal to each other

(adjacent 1 + adjacent 2) tan theta one = adjacent 2 tan theta two

distripute

adjacent 1 tan theta one + adjacent 2 tan theta one = adjacent 2 tan theta two

subtract from both sides

adjacent 1 tan theta one = adjacent 2 tan theta two - adjacent 2 tan theta one

factor or whatever

adjacent 1 tan theta one = adjacent 2 (tan theta two - tan theta one)

muliply by inverse to solve for adjacent 2

(adjacent 1 tan theta one = adjacent 2 (tan theta two - tan theta one)) ((tan theta two - tan theta one)^-1

adjacent two = (tan theta two - tan theta one)^-1 adjacent 1 tan theta one

plug and chug for adjacent two

adjacent two = (tan 32 degrees - tan 40 degrees)^-1 (50 ft) tan theta 40 degrees

I got negative 195.8 feet?

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