Question

In △ABC, m∠A = α, m∠B = β, m∠C = γ. AB = c, AC = b, BC = a. . Find the angles of the triangle knowing that the sides of the triangle are: a=39.5, b=45.2, c=67.1

Answer 1

Answer:

m < A =   34.44 degrees.

m < B =   40.60 degrees.

m < C = 104.96 degrees.

Step-by-step explanation:

Using the Cosine Rule we have;

c^2 = a^2 + b^2 - 2ab cos C

67.1^2 = 45.2^2 + 39.5^2 - 2*39.5*45.2 cos C

cos C =  (67.1^2 - 45.2^2 - 39.5^2) / (-2*39.5*45.2)

cos C = -0.25841

C = 104.96 degrees

Using the Sine Rule:

c / sin C = b / sin B

67.1/ sin 104.96 = 45.2 / sin B

sin B = (45.2 * sin 104.96) / 67.1

= 0.65079

B = 40.60 degrees.

So A = 180 - 40.60 - 104.96 = 34.44 degrees.

Answer 2

Measure of angles A, B and C will be 34.73°, 40.69° and 104.58° respectively.

    Given in ΔABC,

• BC = 39.5, AB = 67.1, AC = 45.2

By applying cosine rule to get the measure of angle A,

c² = a² + b² - 2a.b.cosC

By substituting the values in the expression,

(67.1)² = (39.5)² + (45.2)² - 2(39.5)(45.2)cosC

4502.41 = 1560.25 + 2043.04 - 3570.8(cosC)

899.12 = -3570.8(cosC)

cosC = -0.251798

C = 104.58°

By applying sine rule in triangle ABC,

$\frac{\text{sinA}}{a}=\frac{\text{sinB}}{b}=\frac{\text{sinC}}{c}$

$\frac{\text{sinB}}{45.2}=\frac{\text{sin(104.58)}}{67.1}$

sinB = 0.651929

B = 40.69°

By triangle sum theorem,

m∠A + m∠B + m∠C = 180°

m∠A + 40.69° + 104.58° = 180°

A = 34.73°

   Measure of angles A, B and C will be 34.73°, 40.69° and 104.58° respectively.

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