Across-section of an airplane wing is shown. measurements of the thickness of the wing, in centimeters, at 25-centimeter intervals are 6.1, 19.8, 27.2, 29.0, 27.5, 26.9, 23.7, 20.4, 16.3, 9.1, and 2.9. use the midpoint rule with n = 5 to estimate the area of the wing's cross-section if a = 250. (assume the thickness of the edges is nonzero.)

The thicknesses in centimeters at the midpoints of 5 equal-width intervals are 19.8, 29.0, 26.9, 20.4, and 9.1. The width of each of the 5 intervals is 50 cm.

An estimate of the area of the cross-section using the midpoint rule.

According to the midpoint rule, the area estimate is the sum of the midpoint values multiplied by the width of the interval. That is ...

... Area ≈ (50 cm)×((19.8 +29.0 +26.9 +20.4 +9.1) cm) = 5,260 cm²

we know that

applying the law of sines

°

so

∠j=18°

∠k=29°

∠l=180-(18+29)=133°

the answer part a) is

the measure of angle k is  ≈29°

part b)  the measure of angle k could also be

(180-29)=151°

so

∠j=18°

∠k=151°

∠l=180-(18+151)=11°

the answer part b) is

the measure of angle k could also be 151°

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