Question

A triangle has side lengths of 9 in, 13 in, and 20 in. What is the measurement of this triangle's largest angle?

Answer 1

1. Given any triangle ABC with sides BC=a, AC=b and AB=c, the following are true :

     i) the larger the angle, the larger the side in front of it, and the other way around as well. (Sine Law) Let a=20 in, then the largest angle is angle A.

      ii) Given the measures of the sides of a triangle. Then the cosines of any of the angles can be found by the following formula:

       a^{2}=b ^{2}+c ^{2}-2bc(cosA)

2.

20^{2}=9 ^{2}+13 ^{2}-2*9*13(cosA)

400=81+169-234(cosA)   150=-234(cosA)

cosA=150/-234= -0.641

3. m(A) = Arccos(-0.641)≈130°,

4. Remark: We calculate Arccos with a scientific calculator or computer software unless it is one of the well known values, ex Arccos(0.5)=60°, Arccos(-0.5)=120° etc

Answer 2

Answer:

Measurement of this triangle's largest angle = 129.87°

Step-by-step explanation:

Let a = 9 in, b = 13 in and c = 20 in

We have cosine rule $cosA=\frac{b^2+c^2-a^2}{2bc}$

    $cosA=\frac{13^2+20^2-9^2}{2\times 13\times 20}=0.938\\\\A=20.21^0\\\\cosB=\frac{9^2+20^2-13^2}{2\times 9\times 20}=0.867\\\\B=29.93^0\\\\cosC=\frac{9^2+13^2-20^2}{2\times 9\times 13}=-0.641\\\\C=129.87^0$

Measurement of this triangle's largest angle = 129.87°