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27 March, 17:07

ABC is a triangle. Angle A is 62°, correct to the nearest degree. Angle B is 53.4°, correct to the nearest tenth of a degree. (a) Write down the lower bound for angle B. (b) Calculate the upper bound for angle C.

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  1. 27 March, 17:24
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    a) 53.35

    b) 65.15

    Step-by-step explanation:

    Given triangle ABC.

    Angle A = 62° (to the nearest degree)

    Angle B = 53.4° (to tge the nearest tenth of a degree)

    Since angle A is corrected to the nearest degree, let's assume original figure of angle A is between the range of 61.5 to 62.4

    Thus, 61.5 ≤ 62.4

    Similarly, angle B is corrected to the nearest tenth of a degree, let's assume original figure of angle B was between the range of 53.35 to 53.44

    Thus, 53.35 ≤ 53.44

    a) The lower bound for angle B = 53.35

    b) To calculate the upper bound for angle C, let's add the lower bound figures of the range of angle A and B, then subtract from 180°.

    180 - (61.5 + 53.35) = 65.15

    The upper bound for Angle C is 65.15
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