In ΔABC, ∠A measures 132° and the values of b and a are 36 and 70, respectively. Find the remaining measurements of the triangle, and round your answers to the nearest tenth. ∠B = 82.2°, ∠C = 62.8°, c = 40.6 ∠B = 82.2°, ∠C = 62.8°, c = 30 ∠B = 22.5°, ∠C = 25.5°, c = 40.6 ∠B = 22.5°, ∠C = 25.5°, c = 30

The correct answer is (C) Angle B = 22.5, Angle C = 25.5 and line c = 40.6

Step-by-step explanation: Please refer to the attached diagram for details.

The triangle has been given with angle A measuring 132 degrees and line a measuring 70 units. Angle B is not given but line b measures 36 units. We can apply the sine rule which states that;

a/SinA = b/SinB = c/SinC

Hence, we now have

70/Sin132 = 36/SinB

By cross multiplication we now have

70 (SinB) = 36 (Sin132)

SinB = {36 (Sin132)}/70

SinB = (36 x 0.7432)/70

SinB = 0.3822

By use of calculator or table of values,

B = 22.5 degrees.

To calculate angle C,

Angle C = 180 - (132 + 22.5)

{Sum of angles in a triangle equals 180}

C = 180 - 154.5

C = 25.5 degrees

To calculate line C, having derived angle C,

70/Sin132 = c/Sin25.5

By cross multiplication we now have

70 (Sin25.5) = c (Sin132)

70 x 0.4305 = c x 0.7432

(70 x 0.4305)/0.7432 = c

40.55 = c

c is approximately 40.6

Therefore, Angle B = 22.5, Angle C = 25.5, and line c = 40.6

answer rather than having to add and subtract the measurements of the semi-circle, notice that the semicircle to the left of ad is actually identical in area to the blank semicircle-shaped space to the left of bc. so if we transferred the red semicircle into the blank space, we would end up with a square of side length 24 cm. the area of a square is s^2 = 24^2 = 676 cm^2.

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