Math destroyer Test 5 #27

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SaSSyBruin

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The questions ask to find the length of side BC of the triangle below, given that sin (17.5)= 0.3 and the triangle picture shows another angle being 30 degrees and gives one side of length 100

The triangle is an non-right triangle, so the solution states we should use the LAW OF SINES since Angle-Side-Angle is given.

I'm not sure how the came up with the answer. I understand the initial setup of the law of sines but upon plugging in, I dont understand how they rearranged it to find the answer.

Sidenote: Law of Sines is using when ASA, SSA, or AAS are given..
 
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100/sin(30) = BC/sin(17.5). sin(17.5) = 0.3 ( given). sin(30) = 0.5 ( u should know it)

100/0.5 = BC/sin(17.5)
200 = BC/0.3 then cross multiply
200*0.3=BC
60=BC
 
I understand how the equation is supposed to be setup, but I do not understand how they initial setup rearranges or is setup to find to length of side BC. Thanks for the help, test day soon! Very appreciated.
 
Ohhhhhhhhhhh okay.

You have to separate the "a/sin(A) = b/sin(B) etc" line from the very next line. You're thinking that they took that first line and plugged stuff in and got to the second line.

In reality that first line is the general formula. You only need to use one equality there:

a/sin(A) = b/sin(B)

The terms 'a' and 'b' and 'c' are your side lengths opposite their respective angles 'A', 'B', and 'C'.

So side [BC] is opposite angle A (why? draw the triangle) with side length of 'a'.

With that in mind you can solve the problem. [BC] has absolutely nothing to do with side lengths 'b' and 'c'.
 
It was given in the problem, they provided it for you. There is no way you have to be able to calculate sin(17.5) on the real DAT.
 
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