Well, this is an ambiguous case, we must solve for the third side of the triangle. Frankly I forgot how to do this but I hope this site will help SparkNotes: Solving Oblique Triangles: The Ambiguous Case

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- Mar 15th 2014, 01:24 PM #1

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- Mar 15th 2014, 01:31 PM #2
## Re: SAT Question

Well, this is an ambiguous case, we must solve for the third side of the triangle. Frankly I forgot how to do this but I hope this site will help SparkNotes: Solving Oblique Triangles: The Ambiguous Case

- Mar 15th 2014, 01:36 PM #3

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- Mar 15th 2014, 01:37 PM #4

- Mar 15th 2014, 01:54 PM #5

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- Mar 15th 2014, 03:55 PM #6

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- Mar 15th 2014, 04:03 PM #7

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- Mar 15th 2014, 04:19 PM #8
## Re: SAT Question

It does not have to.

Suppose that the length of $\overline{AB}$ is 10. Think of a semicircle centered at $A$ with radius 7 beginning from the segment.

If $C$ is any point on that semicircle not on $\overleftrightarrow {AB}$ then the max area for $\Delta ABC$ is when $\overline {AB} \bot \overline {AC} $

- Mar 15th 2014, 04:40 PM #9

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- Mar 16th 2014, 12:48 AM #10

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## Re: SAT Question

These tests dont want you to over think but everybody does, in its simplest form of 1/2 * base * height. .5 * 10*7 is 35. We all want to work out the third side as we should but as the angle Theta becomes Obtuse, what happens to the area as the angle widens and the third leg grows longer and the first two legs given spread wider apart, the base will stay the same and the height will grow shorter. Again you must assume that the leg that is growing or shrinking is not the base or height. A right triangle is the greatest Area with two fixed legs of length. Area of a triangle given three sides - Heron's Formula - Math Open Reference, something to play with.