<p>A man walks 300mEast along a river until he reached a mill. then he walked 200m S15E to the post on the edge of a road. Finally he followed the road back to his starting spot... What is the total area?
I don't think the quad. is cyclic (maybe it is...but i don't think so). Right now i don't have a clue in heck, but it's supposed to be law of sines/law of cosines level...
thanks a ton in advance,
zumy</p>
<p>it's an SAT style question...</p>
<p>I am assuming that the road follows a direct path back to his starting point.</p>
<p>You would use the equation Area = (1/2)a<em>b</em>sin(C) to find the answer.
So Area = (1/2)(300)(200)sin(90+15)<br>
(South means 90 degrees from East, and 15 degrees East is added onto that)
so... 30000sin(105) = 28977.77</p>
<p>Very good!
but that's only one of the triangles...the pic looks like this
B
---------------
- - C
A - _ -
- _ -<br>
- D</p>
<p>A=500m, B=300m, C=200m, D?</p>
<p>nvm about the picture, but there's two triangles,sorry</p>
<p>I can't make sense of that picture...sorry.</p>
<p>sorry about the pic...you are actually right the first time, but there's 2 triangles...what you need to do is add that area to the area of the 2nd triangle. this can be found by finding the last side of the Quad. that took me the most time, but once you find that it ends up being apprx. 150060m2.
but thanks again</p>
<p>How did you get that equation?</p>
<p>The way you get the equation is quite ismple, if you understand the basic trigonometric ratios.</p>
<p>Visualize a triangle, with base a, angle left of the base C, angle right of the base B, top angle A, with side b opposite angle B.</p>
<p>You know the formula for the area of a triangle is Area = (1/2)<em>base</em>height.</p>
<p>Now visualize the height of the triangle, from angle A and perpendicular to base a.</p>
<p>Sin(x) = opposite/hypotenuse.
So sin(C) = (height)/(b)
So to get the height, you must multiply sinC by side b. ((h/b)*b=h))</p>
<p>So (1/2)(base<em>height) becomes (1/2)(a</em>(bsinC)).</p>
<p>flipsta is right, its basic trig.</p>
<p>Ohhh, I see. Sort of resembles the law of cosine.</p>