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ryzh [129]
3 years ago
8

A triangular field has sides 218.5 and 213.3 and the angle between them measures 58.96°. Find the area of the field.

Mathematics
1 answer:
Zolol [24]3 years ago
4 0

Answer:

The area of the field = 19966.21 units²

Step-by-step explanation:

* Lets explain how to find area of a triangle by trigonometry rule

- In any triangle if you have the lengths of two sides and the measure

 of the including angle between these two sides, then the area of the

 triangle is A = \frac{1}{2}s_{1}s_{2}sin\alpha , wher α is the

 including angle between them

* Lets solve the problem

∵ The field is shaped triangle

∵ The lengths of two sides of the field are 218.5 and 213.3

∴ s1 = 218.5

∴ s2 = 213.3

∵ The measure of the angle between the two sides is 58.96°

∴ α = 58.96°

- Lets find the area using the rule of trigonometry

∴ A=\frac{1}{2}(218.5)(213.3)sin(58.96)=19966.21

∴ The area of the field = 19966.21 units²

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Use the mid-point rule with n = 4 to approximate the area of the region bounded by y = x3 and y = x. (10 points)
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See the graph attached.

The midpoint rule states that you can calculate the area under a curve by using the formula:
M_{n} = \frac{b - a}{2} [ f(\frac{x_{0} + x_{1} }{2}) +  f(\frac{x_{1} + x_{2} }{2}) + ... +  f(\frac{x_{n-1} + x_{n} }{2})]

In your case:
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Therefore, you'll have:
M_{4} = \frac{1 - 0}{4} [ f(\frac{0 +  \frac{1}{4} }{2}) +  f(\frac{ \frac{1}{4} + \frac{1}{2} }{2}) +  f(\frac{\frac{1}{2} + \frac{3}{4} }{2}) + f(\frac{\frac{3}{4} + 1} {2})]
M_{4} = \frac{1}{4} [ f(\frac{1}{8}) +  f(\frac{3}{8}) +  f(\frac{5}{8}) + f(\frac{7}{8})]

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M_{4} = \frac{1}{4} [(\frac{1}{8} - \frac{1}{512}) + (\frac{3}{8} - \frac{27}{512}) + (\frac{5}{8} - \frac{125}{512}) + (\frac{7}{8} - \frac{343}{512})]

M₄ = 1/4 · (2 - 478/512)
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Hence, the <span>area of the region bounded by y = x³ and y = x</span> is approximately 0.267 square units.

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