I solved the problem can someone check to see if I did it right? Medal award To find a distance across a small lake, a surveyor has taken the measurements shown. Find the distance across the lake, AB
I multiplied sin 40.3 (3.56) then I multiplied both sides by 3.56 to illuminate the x by itself I got x=2.302571615
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I made him log off x'D
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what are you talking about :D
abhisar was iron man :D
i was watching him :D
If i may, you should bump this question now so people see it again LOL
lol ok )
Im hulk... *farts rainbows and flys away*
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@amistre64 Help dis guy -_-
lots of spam in this post ... whats the question?
use law of cosines ...
Welp... sorry for the spam LOL
but is it right? The way I solved the problem?
no, its not, you assumed a right triangle, but its not a right triangle
|dw:1430968367018:dw| \[c^2=a^2+b^2-2ab~\cos(\alpha)\]
how do I know what to put in for the c^2 if i'm only given side length 2.82 and 3.56?
the value of c is what you are trying to find ... its not something you "put in" its something you get as a result.
if you have 2 sides and the angle between them, then you have all you need to calculate the missing side.
so I substitute the 2.82 as the a and the 3.56 as the b into the equation. We are solving for c?
yes, and the angle is subbed into the cosine part then sqrt to solve for c
Rainbows
amistre64, so would you set it up as c^2=(2.82)^2+(3.56)^2-2(2.82)(3.56)Cos c
yes, but the last little bit is off \[ c^2=(2.82)^2+(3.56)^2-2(2.82)(3.56)\cos\color{red}{40.3} \]
2.30496 would be the distance across the lake ab
yes, rounded to whatever decimation you like
ok thanks so much!
youre welcome
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