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Mariulka [41]
3 years ago
6

Easy brainliest, please help me..

Mathematics
1 answer:
DiKsa [7]3 years ago
5 0

Answer:

Solution given:

arcCD=2×35=70°[inscribed angle is half to the central angle]

r=10in

area of segment :\frac{70}{360} π×r²

:\frac{70}{360} π×10²

=\frac{175}{9} π in²

<u>C</u><u>.</u><u>\frac{175}{9} π in²</u><u> </u><u>i</u><u>s</u><u> </u><u>a</u><u> </u><u>right</u><u> </u><u>a</u><u>n</u><u>s</u><u>w</u><u>e</u><u>r</u><u>.</u>

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Pls solve question 8
s2008m [1.1K]

The value of the angle ∠AVB is 53°, side VB = 132.83 m, side OV = 119 m

<h3>How to find the side length from bearings?</h3>

8) From the given triangle we see that;

∠VAO = 37°

∠VBO = 64°

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a) Now, we know that sum of angles in a triangle is 180°. Thus, for triangle AVO, we can say that;

∠VAO + ∠VOA  + ∠AVB = 180°

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b) Using triangle rules we can say that;

tan 64 = VO/BO

Similarly, we can say that;

tan 37 = VO/(100 + BO)

Dividing both equations gives;

tan 64/tan 37 = (100 + BO)/BO

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Using trigonometric ratio;

58.14/VB = cos 64

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OV = 132.83 * sin 64

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Read more about Side lengths from bearings at; brainly.com/question/23427938

#SPJ1

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