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Math Help - Basic Trigonometry

  1. #1
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    Basic Trigonometry

    The triangle ABC has sides AB [] AC [] and BC [], as well as cosBaC []

    a] Write down and simplify a quadratic equation satisfied by and hence evaluate

    b] Express sinBaC in the form

    c] Express sinAcB in the form

    I have no idea how to solve these, it's the most difficult question I have out of the exam papers and even though I (sort of) understand trigonometry at the moment this is just impossible to me.
    Could anyone please help?

    For a] I have a value of 8, and this I know to be right

    For b] I have \frac{sin73.4}{9} = \frac{sinB}{8} but I don't know how you can express it in the form asked for.

    Please could someone help?
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  2. #2
    Junior Member JoanF's Avatar
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    Quote Originally Posted by db5vry View Post
    The triangle ABC has sides AB [] AC [] and BC [], as well as cosBaC []

    a] Write down and simplify a quadratic equation satisfied by and hence evaluate

    b] Express sinBaC in the form

    c] Express sinAcB in the form

    I have no idea how to solve these, it's the most difficult question I have out of the exam papers and even though I (sort of) understand trigonometry at the moment this is just impossible to me.
    Could anyone please help?

    For a] I have a value of 8, and this I know to be right

    For b] I have \frac{sin73.4}{9} = \frac{sinB}{8} but I don't know how you can express it in the form asked for.

    Please could someone help?

    b) you can use the formula: sin^2 A + cos^2 A =1

    c) in b) you will obtain sin BaC= (square of 45)/7. you know 7cm is one side of the triangle so you know that (square of 45) cm is the height. so sin AcB=(square of 45)/9 [because 9 is the other side
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  3. #3
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    Hello, db5vry!

    The triangle ABC has sides: AB = 7,\;BC = 9,\;AC = x and \cos A = \tfrac{2}{7}
    Code:
                B
                *
               *  *
           7  *     *   9
             *        *
            *           *
           *              *
        A *  *  *  *  *  *  * C
                  x
    (a) Write down and simplify a quadratic equation satisfied by x and evaluate x

    Law of Cosines: . a^2 \:=\:b^2 + c^2 - 2bc\cos A

    We have: . 9^2 \:=\:x^2 + 7^2-2(x)(7)\cos A \quad\Rightarrow\quad 81 \:=\:x^2 + 49 - 14x\left(\tfrac{2}{7}\right)

    . . which simplifies to: . \boxed{x^2 - 4x - 32 \:=\:0}\quad\Rightarrow\quad (x-8)(x+4) \:=\:0


    Therefore: . x \,=\,8\quad {\color{red}\rlap{///////}}x \,=\,-4




    (b) Express \sin B in the form \frac{\sqrt{m}}{n}

    Law of Sines: . \frac{\sin B}{b} \:=\:\frac{\sin A}{a} \quad\Rightarrow\quad \sin B \:=\:\frac{b\sin A}{a}\;\;{\color{blue}[1]}

    We have: . b = x = 8,\:a = 9
    And: . \sin^2\!A \:=\:1-\cos^2\!A \:=\:1-\left(\tfrac{2}{7}\right)^2 \:=\:1 - \frac{45}{49} \quad\Rightarrow\quad \sin A \:=\:\frac{\sqrt{45}}{7}

    Substitute into [1]: . \sin B \:=\:\frac{8\left(\frac{\sqrt{45}}{7}\right)}{9} \:=\:\frac{8\sqrt{45}}{63}  \:=\:\frac{\sqrt{2880}}{63}




    (c] Express \sin C in the form \frac{\sqrt{P}}{3}

    Law of Sines: . \frac{\sin C}{c} \:=\:\frac{\sin B}{b} \quad\Rightarrow\quad \sin C \:=\:\frac{c\sin B}{b}\;\;{\color{blue}[2]}

    We have: . c = 7,\;b = x = 8,\;\sin B = \frac{8\sqrt{5}}{21}

    Substitute into [2]: . \sin C \:=\:\frac{7\left(\frac{8\sqrt{5}}{21}\right)}{8} \:=\;\frac{\sqrt{5}}{3}

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