<p>The area of a triangle with sides 3,5,7 is
a) 7.5
b) 6.5
c) 3.75
d) 13.0
e)2.4
I totally forgot how to do this problem. I know it has to do with law of cosines but I still have no idea.</p>
<p>The polar coordinates of a point P are (2,240degrees). The Cartesian (rectangular) coordinates of P are
A) -1, -rad 3
B) -1, rad 3
C) -rad 3, -1
D) -rad 3, 1
e) none of the above
No idea what polar coordinates are or how to solve this problem</p>
<ol>
<li>c^2 = a^2 + b^2 - 2abcosC
49 = 9 + 25 - 30cosC
15 = -30cosC
-.5 = cosC
C = 120°</li>
</ol>
<p>k = 1/2absinC
k = 1/2(3)(5)(sin120°)
k = 6.5, choice B</p>
<p>u could use hero's formula for q1, ie, root of s(s-a)(s-b)(s-c) where s is semiperimeter and a,b,c are the sides. you get root of 42.1875=6.5 approx.</p>
<p>the answer to question 2 is A.</p>
<p>polar coordinates are written in the form (r,theta).</p>
<p>To convert polar form to rectangular form you use this equation</p>
<p>x = rcos(theta)
y = rsin(theta)</p>
<p>ooooh yeahhh, forgot a bout heron's formula.. that's probably a lot more efficient!</p>
<p>these two questions can be solved within 10 seconds or less each based on how fast your fingers can move.</p>
<p>wanna know how?</p>
<p>alright, for the first one
The area of a triangle with sides 3,5,7 is
a) 7.5
b) 6.5
c) 3.75
d) 13.0
e)2.4</p>
<p>Use the herons formula function, type in hero(3,5,7) and the answer will be 6.49 (B)</p>
<p>next one
The polar coordinates of a point P are (2,240 degrees). The Cartesian (rectangular) coordinates of P are
A) -1, -rad 3
B) -1, rad 3
C) -rad 3, -1
D) -rad 3, 1
e) none of the above </p>
<p>Using the polar to rectangular converter program you can find that the answer is (-1, -1.73)
I don't know what the "rad" in your problem stands for...so I would go with E, but i could be wrong.</p>
<p>PS; Programs/functions for ti-89 are here...<a href="http://www.satprogs.com%5B/url%5D">www.satprogs.com</a> take a look</p>
<p>"rad" must stand for radical. Makes A) the right answer.</p>
<p>oh true that...thanks gcf</p>