Question

In △ABC,a=13, b=21, and c=27. Find m∠A. A. 18.4 B. 31.5 C. 28.0 D. 103.0

Answer 1

Answer:

Option C (28.0°)

Step-by-step explanation:

The questions which involve calculating the angles and the sides of a triangle either require the sine rule or the cosine rule. In this question, the three sides are given and one unknown angle has to be calculated. Therefore, cosine rule will be used. The cosine rule is:

a^2 = b^2 + c^2 - 2*b*c*cos(A°).

The question specifies that a=13, b=21, and c=27. Plugging in the values:

13^2 = 21^2 + 27^2 - 2(21)(27)*cos(A°).

Simplifying gives:

-1001 = -1134*cos(A°)

Isolating cos(A°) gives:

cos(A°) = 0.88271604938

Taking cosine inverse on the both sides gives:

A° = arccos(0.88271604938). Therefore, using a calculator, A° = 28.0 (correct to one decimal place).

This means that the Option C is the correct choice!!!

Answer 2

For this case we have that by definition, the cosine theorem states that:

$a ^ 2 = b ^ 2 + c ^ 2-2bc * Cos (A)$

According to the data we have:

$a = 13\\b = 21\\c = 27$

Substituting we have:

$13 ^ 2 = 21 ^ 2 + 27 ^ 2-2 (21) (27) * Cos (A)\\169 = 441 + 729-1134 * Cos (A)\\169 = 1170-1134 * Cos (A)\\169-1170 = -1134 * Cos (A)\\-1001 = -1134 * Cos (A)\\Cos (A) = \frac {1001} {1134}\\Cos (A) = 0.8827\\A = arc cos (0.8827)\\A = 28.03$

Answer:

Option C