<p>Hey guys... what's up?</p>
<p>I need help solving a problem, so if you'd be so kind...</p>
<p>This comes from Barron's (pg. 33, Example 5)</p>
<p>Find the acute angle formed by lines 2x + 3y + 5 = 0 and 3x - 5y + 8 = 0.</p>
<p>Seems kinda tricky, any help is appreciated.</p>
<p>There is a formula for this mentioned in the review chapter. Simply plug the numbers into the formlua. It is a good idea to have this formlua in you calculator (see the section on programs, this one is listed)</p>
<p>a formula? really? do you know where... cuz I can't seem to find it.</p>
<p>oh haha nvm. got it. thanks mate.</p>
<p>yeah... its called using tangents. :P once u find the slopes then use tri to find the angles. very easy</p>
<p>wait its not on my barron's workbook......................</p>
<p>well...i'm not sure if I computed right but my answer is something around 64,65 degrees..hmm...
my method:
slope1= -2/3
slope2=3/5
tan(x)=abs[(slope1-slope2)/(1+slope1*slope2)]
tan(x)=19/9....etc...you do the rest of the computing ...but it's something like this:)</p>
<p>There is a much easier way, possibly using vectors, but I don't remember it. Sorry.</p>
<p>well this one isn't as complicated as it seems...you just have to know the slope formula and the one for tan(x) then have a calculator to find x...that's it;)</p>
<p>You can also just use slopes for tan on each real fast:
arctan(3/5) - arctan(-2/3) = 64.654
never seen that formula above, I personally would steer away from that kind of thing.</p>
<p>Whats the slope formula? y2 - y1/ x2-x1?</p>
<p>how do you enter arctan into your calculator?</p>
<p>there should be a button for Tan^-1.. In my TI-83 its, 2nd+Tan..</p>
<p>its the tan-1 button</p>
<p>isn't the vector way the dot product of two vectors multiplied by the sin of the interior angle... sheesh... i think thats area. i told you i stink at math</p>
<p>line 1: 2x+3y=-5
line 2: 3x-5y=-8</p>
<p>Find the direction vector of each line:
For line 1: A=(2,3)
line 2: B=(3,-5)</p>
<p>Using dot product:
cos(theta)=AdotB/((magA)(magB))
cos(theta)=(2<em>3+3</em>-5)/((sqrt(4+9))(sqrt(9+25)))
cos(theta)=-9/((sqrt(13))(sqrt(34)))
cos(theta)=-9/(sqrt(442))
theta= about 64.65 degrees</p>
<p>PoloniusMouth, I think you're confusing dot product with cross product.</p>
<p>i know... its been years since algebra 2. I completely forgot about vectors and stuff. Actually, I never liked it to begin with. :S</p>
<p>I think I'll have to go over dot products. Probably some integrals too b4 school starts</p>
<p>Thanks towerpumpkin, that's what I meant earlier. Less than a month into summer and I'm already forgetting everything. Well actually, its been half a year since I've had math...yay block scheduling?</p>
<p>is it bad that i just finished precalc and this entire post seems like another language? we never went over vectors or that formula or anything similar to this haha</p>