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The answer is 3 and 5. Explanation?enter image description here

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    $\begingroup$ This is a triangle with three knowns which are in counter-clockwise order: Side-Side-Angle. You should know that there are generally two possibilities (either in this case $\angle ABC >\frac{\pi}{2}$ or $\angle ABC<\frac{\pi}{2}$). Check any elementary textbook on trigonometry or "algebra 2" (highschool algebra, not abstract algebra) or websites such as this for how to solve for the remaining unknowns of a triangle. $\endgroup$
    – JMoravitz
    Sep 12, 2015 at 18:44

3 Answers 3

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Let $BC=x$. By the Law of Cosines:

$$7^2=8^2+x^2-2\cdot 8\cdot x\cdot \cos\frac{\pi}{3}$$

$\cos\frac{\pi}{3}=\frac{1}{2}$, so

$$\iff x^2-8x+15=0\iff (x-4)^2=1$$

$$\iff x-4=\pm 1\iff x\in\{3,5\}$$

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For SAT2 math, you ought to memorize both the Law of Cosines and the Law of Sines. They're both useful to rapidly solve problems with non-right triangles. This one uses the Law of Cosines, where $a,b,c$ are sides of a triangle with opposing angles $A,B,C$.

$$c^2 = a^2 + b^2 -2ac\cos C$$

In your case,

$$7^2 = 8^2 + x^2 -2(8)(x)\cos \frac{\pi}{3}$$

You can either solve this as a quadratic equation (preferably) or substitute your answer choices to find the solution.

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Draw perpendicular from point B to AC, you will get two right angle triangles.

I will leave rest of the work to consider self.

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