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    • Thread Starter

    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 .
    • TSR Group Staff

    TSR Group Staff
    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]
    • TSR Group Staff

    TSR Group Staff
    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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Updated: March 4, 2006

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