A question from ACT math:
I'm wondering except using the law of cosine to get the answer K, is there any faster way to figure it out?
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1Hint : If $c$ is the largest side, we must have $a^2+b^2<c^2$ – Peter Dec 08 '20 at 13:58
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In this case, J is also a obtuse triangle? – HypnoticBuggyWraithVirileBevy Dec 08 '20 at 14:01
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You had the right idea. The cosine law is in fact the key. The numerator is negative if and only if $a^2+b^2<c^2$ in which case we get a negative cosine leading to an angle exceeding $90$° – Peter Dec 08 '20 at 14:03
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$7^2+10^2=149>144=12^2$ , so no – Peter Dec 08 '20 at 14:04
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Don't forget to show that we have a triangle with the given sides at all, which is however easy to check. – Peter Dec 08 '20 at 14:06
1 Answers
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If $\angle A$ is an obtuse angle in a triangle.
$$90<\angle A<180, -1<\cos\angle A<0$$
$$\cos\angle A= \frac{b^2+c^2-a^2}{2bc} <0$$
using only numerator since denominator is always positive
$$a^2>b^2+c^2$$
where $a$ is the longest side
$$16^2>11^2+8^2$$
Lion Heart
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This is exactly how we get the criterion, but $0<cos(\alpha)<-1$ is the wrong double- inequality. – Peter Dec 08 '20 at 14:15