Question

In △ABC,a=11, b=20, and c=28. Find m∠A.

Answer 1

Answer:

About 18.36 degrees.

Step-by-step explanation:

We are going to use law of cosines since we are given the three side measurements.

$a^2=b^2+c^2-2bc \cos(A)$

$11^2=20^2+28^2-2(20)(28) \cos(A)$

$121=400+784-40(28) \cos(A)$

$121=1184-1120 \cos(A)$

Subtract 1184 on both sides:

$121-1184=-1120 \cos(A)$

$-1063=-1120 \cos(A)$

Divide both sides by -1120:

$\frac{-1063}{-1120}=\cos(A)$

$\frac{1063}{1120}=\cos(A)$

Take cosine's inverse on both sides:

$\cos^{-1}(\frac{1063}{1120})=A$

Put left side into calculator:

$18.36^\circ \approx A$

Answer 2

Measure of angle A is  m∠A = $18.358^{0}$.

What is law of cosines?

The law of cosine states that the square of any one side of a triangle is equal to the difference between the sum of squares of the other two sides and double the product of other sides and cosine angle included between them.

Let a, b, and c be the lengths of the three sides of a triangle and A, B, and C be the three angles of the triangle. Then, the law of cosine states that: $a^{2} =b^{2} +c^{2} -2bccosA$

Given

a = 11, b = 20, c = 28

Substitute c = 28 into $\overline A \overline B$ = c :  $\overline A \overline B$ = 28

Substitute a = 11 into $\overline B \overline C$ = a :  $\overline B \overline C$ = 11

Substitute b = 11 into $\overline A \overline C$ = b :  $\overline A \overline C$ = 11

Law of cosines:

$\overline B \overline C^{2} = \overline A \overline C^{2} +\overline A \overline B^{2} -2\overline A \overline C \times \overline A \overline B \times cos( < BAC)$

$11^{2} = 20^{2} +28^{2} -2 \times 20 \times 28 \times cos( < BAC)$

$cos( < BAC) = 11^{2}-20^{2} -28^{2} +2 \times 20 \times 28$

$cos( < BAC) = 57$

$< BAC = cos^{-1} (57)$

< BAC = $18.358^{0}$

Measure of angle A is  m∠A = $18.358^{0}$.

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