Triangle Side Lengths: Find the Third Side with 10 and 14 Units & 30° Angle

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To find the third side of a triangle with sides measuring 10 and 14 units and a non-included angle of 30 degrees, the Law of Cosines is applicable. The relevant formula states that c² = a² + b² - 2ab * cos(C), where C is the angle between sides a and b. Users are encouraged to assign variables to the sides and angles for clarity. Additionally, sketching the triangle can aid in visualizing the problem. The discussion highlights the importance of proper forum placement for homework-related queries.
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A triangle has sides of length 10 and 14 units respectively, and one of the non-included angles is 30 degrees. Find the possible lengths of the third side. The answers is the product of these lengths.
 
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Phanatic 12 said:
A triangle has sides of length 10 and 14 units respectively, and one of the non-included angles is 30 degrees. Find the possible lengths of the third side. The answers is the product of these lengths.
This looks like a homework problem, and so should have been posted in the Homework section, in the Precalculus subsection.

What are the relevant formulas? What have you tried?
 


Phanatic 12 said:
A triangle has sides of length 10 and 14 units respectively, and one of the non-included angles is 30 degrees. Find the possible lengths of the third side. The answers is the product of these lengths.
What have you tried so far? For example, do always sketch a triangle so you can see what you're working with.

I'll give you one hint to get you started. First, assign variables to your sides and angles. Say, \angle{A}, \angle{B} and \angle{C} for the angles and a, b and c for the sides. Now, here's something to ponder:
\frac{ \sin{\angle{A}}}{a} = \frac{\sin{\angle{B}}}{b}

EDIT: Oops, I didn't notice he posted it in the wrong forum. My bad.
 
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