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Math Help - three trigo problems

  1. #1
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    three trigo problems

    three trigo problems
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  2. #2
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    Hello, afeasfaerw232312331

    22) Prove that in \Delta ABC:

    (a) \cot\frac{A}{2} \:=\:\tan\left(\frac{B}{2}+\frac{C}{2}\right)\text  {, by using the identity: }\cot x \:=\:\tan(90^o-x)

    (b) Hence show that: . \cot\frac{A}{2} + \cot\frac{B}{2} + \cot\frac{C}{2} \;=\;\cot\frac{A}{2}\cdot\cot\frac{B}{2}\cdot\cot\  frac{C}{2}
    You solved part (a) . . . Nice work!


    (b) We have: . \cot\frac{A}{2} \;\;=\:\;\tan\left(\frac{B}{2} \:+ \:\frac{C}{2}\right) \quad\Rightarrow\quad\frac{1}{\tan\frac{A}{2}} \;\;=\;\;\frac{\tan\frac{B}{2} \:+ \:\tan\frac{C}{2}}{1-\tan\frac{B}{2}\cdot\tan\frac{C}{2}}

    . . . . 1 \:- \:\tan\frac{B}{2}\cdot\tan\frac{C}{2} \;\;=\;\;\tan\frac{A}{2}\cdot\tan\frac{B}{2} \:+ \:\tan\frac{A}{2}\cdot\tan\frac{C}{2}

    . . . . \tan\frac{B}{2}\cdot\tan\frac{C}{2} \:+ \:\tan\frac{A}{2}\cdot\tan\frac{C}{2} \:+ \:\tan\frac{A}{2}\cdot\tan\frac{B}{2} \;\;=\;\;1

    . . . . \frac{1}{\cot\frac{B}{2}\cdot\cot\frac{C}{2}} \:+ \:\frac{1}{\cot\frac{A}{2}\cdot\cot\frac{C}{2}} \:+ \:\frac{1}{\cot\frac{A}{2}\cdot\cot\frac{B}{2}} \;\;=\;\;1


    Multiply through by: . \cot\frac{A}{2}\cdot\cot\frac{B}{2}\cdot\cot\frac{  C}{2}

    . . . . \cot\frac{A}{2} \:+ \:\cot\frac{B}{2} \:+ \:\cot\frac{C}{2} \;\;=\;\;\cot\frac{A}{2}\cdot\cot\frac{B}{2}\cdot\  cot\frac{C}{2}

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  3. #3
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    Hello again, afeasfaerw23231233!

    26. \Delta ABC is equilateral with AB = a.

    M is a point outside the triangle such that: . \angle AMB = 20^o,\;\angle AMC = 30^o,\;\angle BAM = \alpha.

    (a)(i) By considering \Delta ABM, express AM in terms of a\text{ and }\alpha.
    Your answer is correct, but I would write it this way:

    . AM \:=\:\frac{a\cdot\sin[180^o -(20^o+\alpha)]}{\sin20^o} \quad\Rightarrow\quad AM\;=\; \frac{a\cdot\sin(20+\alpha)}{\sin20^o} .[1]




    (a)(ii) By considering \Delta ACM, express AM in terms of a\text{ and }\alpha.

    Since \angle = 60^o,\;\angle MAC \:=\:60^o-\alpha
    .Then: . \angle MCA \:=\:180^o - 30^o - (60^o-\alpha) \;=\;90^o + \alpha

    We have: . \frac{AM}{\sin(90+\alpha)} \:=\:\frac{a}{\sin30^o} \quad\Rightarrow\quad AM \;=\;\frac{a\cdot\sin(90^o+\alpha)}{\frac{1}{2}}

    . . AM \;=\;2a\cdot\sin(90^o+\alpha)\quad\Rightarrow\quad AM \;=\;2a\cdot\cos\alpha .[2]




    (b) Hence, find \alpha.
    Equate [1] and [2]: . \frac{a\cdot\sin(20^o + \alpha)}{\sin20^o} \;=\;2a\cdot\cos\alpha

    . . . . . . \sin(20^o+\alpha) \;=\;2\cdot\sin20^o\cos\alpha

    . . \overbrace{\sin20^o\cos\alpha + \sin\alpha\cos20^o} \;=\;2\cdot\sin20^o\cos\alpha


    . . \underbrace{\sin\alpha\cos20^o - \sin20^o\cos\alpha} \;=\;0

    . . . . . . \sin(\alpha-20^o) \;=\;0

    . . . . . . . \alpha - 20^o \;=\;0^o

    . . . . . . . . \alpha \:=\:20^o

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  4. #4
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    i thought that the the diagram was a 3D pyramid-like solid! so i spent a great deal of time and could not do [a][ii]. now i know it is a 2D diagram. thanks
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