Radian Measure and its applications

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insparato
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Triangle ABC is such that AB = 5cm, AC = 10cm and angle ABC = 90 degrees. An arc of the circle , centre A and radius 5cm cuts ac at D.

a) State, in radians the value of angle BAC
b) Calculate the area of the region enclosed by BC, DC and the arc BD

Thanks alot .
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Dez
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a) First off, we need length BC (which I'm just going to call a -- opposite of angle A)

a2 + c2 = b2 [irritating letter arrangement :P]
a2 + 25 = 100
a2 = 75
a = sqrt(75) or 5sqrt(3)

Now we use the sine rule:

a / sin A = b / sin B
5sqrt(3) / sin A = 10 / sin 90
5sqrt(3) = 10sin A
sqrt(3) / 2 = sin A

A = pi / 3 or 60 degrees [this is one of the so-called famous triangles]
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Dez
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b) Now this gets tricky. You need the formula for an area of a sector = r2T / 2, where T is the angle in radians.

Okay, we know that angle now, and we know the radius. Sooo...

Asec = 25(pi/3) / 2

So A = (5sqrt(3)*5 / 2) - (25(pi/3) / 2)

Let's disregard that denominator for the moment
= 25sqrt(3) - 25(pi/3)
= 25(sqrt(3) - pi/3)

A = {25(sqrt(3) - pi/3)} / 2

You can shove this in a calculator if you like. Will probably produce some horrible decimal. Or you might be able to cancel it down a bit, but that gives potential for error, assuming I haven't already screwed something up along the way
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Moulik
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We need the diagram for the question
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dextrous63
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(Original post by Moulik)
We need the diagram for the question
Since the thread is 14 years old I suspect that the author of the thread might not still be around too often.
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