The area of a triangle is $60$ square inches. Find the length of the side included between $A = 25°$ and $C = 110°$. (Round your answer to one decimal place.)


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  • $\begingroup$ Welcome to Mathematics Stack Exchange! I would suggest you to explain a little bit how you tried to solve it, so other people could help you better. Good luck! $\endgroup$ – iadvd Jul 7 '15 at 0:17
  • 1
    $\begingroup$ Hint: $$\Delta = \frac{b^2}{2(\cot A +\cot C)}$$ $\endgroup$ – Sawarnik Jul 7 '15 at 1:22

hint: $b^2 = a^2+c^2-2ac\cos(45^{\circ})$, and $\dfrac{b}{\sin 45^{\circ}}= \dfrac{a}{\sin 25^{\circ}}= \dfrac{c}{\sin 110^{\circ}}$, can you find $b$. Note that you have: $ac = \dfrac{2S}{\sin 45^{\circ}}=\dfrac{2\cdot 60}{\dfrac{\sqrt{2}}{2}}=120\sqrt{2}$

  • $\begingroup$ Thanks! I now see the sin45 in the problem, and how to plug in ac=120√2, but I'm still confused on how to find (a^2+c^2) for the Law of Cosines formula. I'm sure it's right in front of me. $\endgroup$ – Nathan Jul 7 '15 at 0:45
  • $\begingroup$ @Nathan, use the law of Sines in the second line of the above answer to write $a$ and then $c$ in terms of $b.$ $\endgroup$ – John Molokach Jul 7 '15 at 0:53

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