# Given only angles and area of triangle, find side length. [closed]

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.)

## closed as off-topic by user147263, user99914, Paramanand Singh, drhab, Lee MosherJul 7 '15 at 14:13

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Community, Paramanand Singh, drhab, Lee Mosher
If this question can be reworded to fit the rules in the help center, please edit the question.

• 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! – iadvd Jul 7 '15 at 0:17
• Hint: $$\Delta = \frac{b^2}{2(\cot A +\cot C)}$$ – Sawarnik Jul 7 '15 at 1:22

## 1 Answer

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: $ac = \dfrac{2S}{\sin 45^{\circ}}=\dfrac{2\cdot 60}{\dfrac{\sqrt{2}}{2}}=120\sqrt{2}$

• 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. – Nathan Jul 7 '15 at 0:45
• @Nathan, use the law of Sines in the second line of the above answer to write $a$ and then $c$ in terms of $b.$ – John Molokach Jul 7 '15 at 0:53