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# Sin,Cos,Tan

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How to you find the angle of a right triangle using only side lengths

Jun 5, 2017

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Use the law of cosines to find the missing angle measures. I'll use a picture to illustrate the law of cosines. I think is makes it easier to understand:

Let's say that the lowercase letters have lengths. I'll arbitrarily assign them for you:

a=10

b=15

c=19

Using this information, simply substitute into the law of cosines formula. You will need a calculator that can calculate trigonometric functions:

 $$c^2=a^2+b^2-2ab\cos C$$ Simply plug in the known values and solve for the missing one. $$19^2=10^2+15^2-2(10)(15)\cos C$$ Because you can use a calculator, I would not simplify anything yet. Subtract 10^2+15^2 on both sides. $$19^2-10^2-15^2=-2(10)(15)\cos C$$ Divide by -2(10)(15) on both sides $$\frac{19^2-10^2-15^2}{-2(10)(15)}=\cos C$$ Use the inverse cosine to isolate C. $$C=\cos^{-1}(\frac{19^2-10^2-15^2}{-2(10)(15)})$$ Use your calculator to evaluate this monstrosity $$C\approx96.89^\circ$$ Repeat this process for the other missing angle measures.

One last warning before you go! Be sure that your calculator is on degree mode when doing the final calculation. Otherwise, your answer will be represented differently. In radian mode, $$\cos^{-1}(\frac{19^2-10^2-15^2}{-2(10)(15)})\approx1.69$$. It is not wrong but a triangle's angle measure is usually represented in degrees--not radians. If your calculator can only preform a calculation like this in radian mode, multiply your answer by $$\frac{180}{\pi}$$ to convert.

Jun 5, 2017
edited by TheXSquaredFactor  Jun 5, 2017