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Math Help - need help for basic trigo question

  1. #1
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    need help for basic trigo question

    hi emm i am a begginer new to trigo just learnt it and i need help on this question.
    i have included a basic diagram drawn by me.(sry if the diagram is not very nice)
    i am stuck with the first part already.2nd part i know how do.
    In the diagram,TA and TB are tangents to the circle,centre O at A and B respectively.The radius of the circle is 5cm and AB is 9cm.Find
    a)angle AOB
    b)the perpendicular distance from O to AB
    c)the length of tangent TA
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  2. #2
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    Hey, thats a nice clear diagram. You will need some basic circle geometry to solve these questions.

    First draw a line from point O to T. Where it intersects the line AB I will call point P.

    Now using circle geometry, we know line OP bisects the line AB at 90 degrees (I don't remember the exact wording of the theorem but look it up if you need to).

    So you have a right angled triangle OPA with right angle OPA, hypothenuse 5, and height (9/2) = 4.5

    Qa:
    Let angle OAB = x. Cosine (x) = (adjacent side)/hypothenuse = 4.5/5
    so x = arcsin (4.5/5)

    Qb:
    You actually don't need to use trigonometry here. Pythagoras equation  a^2 + b^2 = c^2 will work with a = PA and c = OA.
    But if you want to use trig, from the previous working, you have angle OAB = x.
    Sine (x) = (opposite side) / hypothenuse. You have hypothenuse = OA = 5, sine (x) you can calculate using the value of x you got from Qa. Now you can solve for opposite side which is perp distance O to AB.

    Qc:
    See if you can do this yourself, ask for help if you need it. Note that the tangent line is always perpendicular to the centre of the circle (i.e. right angles)
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  3. #3
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    Hello, tempq1!

    In the diagram, TA and TB are tangents to the circle O at A and B.
    The radius of the circle is 5 cm and AB is 9 cm.

    Find: a) angle AOB

    Draw OT, intersecting AB at P.
    Let \theta = \angle AOP

    We have: . AP = \frac{9}{2},\;AO = 5

    Hence: . \sin\theta \:=\:\frac{\frac{9}{2}}{5} \;=\;\frac{9}{10}

    . . \theta \;=\;\sin^{-1}(\frac{9}{10}) \;\approx 64.16^o

    Therefore: . \angle AOB \;=\;2\theta \;\approx\;128.3^o




    b) the perpendicular distance from O to AB

    In right triangle APO\!:\;\;OP^2 + (\frac{9}{2})^2 \:=\:5^2 \quad\Rightarrow\quad OP^2 \:=\:\frac{19}{4}

    Therefore: . OP \;=\;\dfrac{\sqrt{19}}{2} cm




    c) the length of tangent TA

    In right triangle OAT\!:\;\;\tan\theta \:=\:\dfrac{T\!A}{5} \quad\Rightarrow\quad T\!A \:=\:5\tan\theta .[1]


    We know that: . \sin\theta \:=\:\frac{9}{10} \:=\:\frac{opp}{hyp}

    \theta is in a right triangle with: . opp = 9,\;hyp = 10
    Pythagorus says: . adj = \sqrt{19}
    . . Hence: . \tan\theta \:=\:\frac{opp}{adj} \:=\:\frac{9}{\sqrt{19}}

    Substitute into [1]: . T\!A \;=\;5\left(\frac{9}{\sqrt{19}}\right) \;=\; \dfrac{45}{\sqrt{19}} cm.

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