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sergeinik [125]
3 years ago
14

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.
Mathematics
1 answer:
inna [77]3 years ago
8 0

Answer:

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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