In $\Delta$ $KLM$, $KL=20$ $LM=13$ m$\angle K$$=40$. What is the range for angle $M$'s measure? Something like between $90^{\circ}$ and $180^{\circ}$.
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I think it's safe enough to say angle $M$ is in the range between $0$ and $180$ degrees. Actually there are precisely two possible values, of which one is less than $90$ degrees and the other is greater than $90$ degrees. This question is far too vague in its current wording. – David K Jan 18 '19 at 02:46
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You can use the law of sine, then find the value exact of angle measure
$$\frac{\sin{(M)}}{20}=\frac{\sin{(K)}}{13}$$
AsdrubalBeltran
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When the given condition is in 'ASS' situation, there could be TWO possible answers for M. See my answer in problem no. 690645. In your case, YES there are two. – Mick May 05 '14 at 00:37
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Drop the altitude $LS$ of the isoceles triangle $\triangle M_1LM_2$.
Then $LS=13\sin\phi$. On the other hand, since $\sin K=\frac{LS}{KL}$ we have $\sin 40^{\circ}=\frac{13\sin\phi}{20}$. Hence, $\sin\phi=\frac{20\sin40^{\circ}}{13}\approx 0.9889$ and $\phi\approx 81.4568^{\circ}$ (the acute angle solution).
The obtuse angle solution is $180^{\circ}-\phi\approx 98.5432^{\circ}.$
Bob Dobbs
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