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Math Help - In a triangle ABC, value of angle C?

  1. #1
    Super Member fardeen_gen's Avatar
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    In a triangle ABC, value of angle C?

    In a triangle ABC, a = 2, b = 3 and tan A = √(3/5) then the value(s) of C are:

    A)√(10)
    B){√(10)}/2
    C){3√(10)}/2
    D){√(10)}/3
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  2. #2
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    Quote Originally Posted by fardeen_gen View Post
    In a triangle ABC, a = 2, b = 3 and tan A = √(3/5) then the value(s) of C are:

    A)√(10)
    B){√(10)}/2
    C){3√(10)}/2
    D){√(10)}/3
    Did you mean get the value(s) of c? If your notation is to be consistent, C represents an angle and c represents a side length .....

    Sine rule: \frac{\sin A}{2} = \frac{\sin B}{3} \Rightarrow \sin B = \frac{3 \sin A}{2}.

    Get the value of \sin A using the given data \tan A = \frac{\sqrt{3}}{\sqrt{5}}. Then solve for B.

    C = 180 - A - B.

    Cosine rule: c^2 = 2^2 + 3^2 - 2(2)(3) \cos C \, ....
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  3. #3
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    Hello, fardeen_gen!

    In a triangle ABC: . a = 2,\;b = 3,\; \tan A = \sqrt{\frac{3}{5}}
    then the value of {\color{red}c} is:

    A)\;\sqrt{10} \qquad B)\;\frac{\sqrt{10}}{2} \qquad C)\;\frac{3\sqrt{10}}{2}\qquad D)\; \frac{\sqrt{10}}{3}

    Since \tan A \:=\:\frac{\sqrt{3}}{\sqrt{5}} \:=\:\frac{opp}{adj} \quad\Rightarrow\quad \text{then: }\:hyp \:=\:\sqrt{8} \:=\:2\sqrt{2}

    . . Hence: . \begin{array}{ccccccc}\sin A \:=\:  \dfrac{\sqrt{3}}{2\sqrt{2}} &\Rightarrow & \boxed{\sin A \:=\:\dfrac{\sqrt{6}}{4}} \\ \\[-3mm]\cos A \:=\:  \dfrac{\sqrt{5}}{2\sqrt{2}} &\Rightarrow& \boxed{\cos A \:=\: \dfrac{\sqrt{10}}{4}} \end{array}



    Law of Sines: . \frac{\sin B}{b} \:=\:\frac{\sin A}{a} \quad\Rightarrow\quad \sin B \:=\:\frac{b\!\cdot\!\sin A}{a}

    . . Hence: . \sin B \:=\:\dfrac{3\cdot\frac{\sqrt{6}}{4}}{2} \quad\Rightarrow\quad \begin{array}{ccc}\boxed{\sin B \:=\: \dfrac{3\sqrt{6}}{8}} \\ \\[-3mm] \boxed{\cos B \:=\: \dfrac{\sqrt{10}}{8}} \end{array}



    Since: . C \:=\:180^o - (A + B),
    . . we have: . \sin C \;=\;\sin\left[180^o-(A + B)\right] \:=\:\sin(A + B)

    \sin C \;=\;\sin A\cos B + \sin B\cos A \;=\;\left(\frac{\sqrt{6}}{4}\right)\left(\frac{\s  qrt{10}}{8}\right) + \left(\frac{3\sqrt{6}}{8}\right)\left(\frac{\sqrt{  10}}{4}\right)

    . . . . = \;\frac{\sqrt{60}}{32} + \frac{3\sqrt{60}}{32} \;=\;\frac{4\sqrt{60}}{32} \;=\;\frac{\sqrt{60}}{8}  \quad \Rightarrow\quad\boxed{\sin C \:=\:\frac{\sqrt{15}}{4}}


    Law of Sines: . \frac{c}{\sin C} \:=\:\frac{a}{\sin A} \quad\Rightarrow\quad c \:=\:\frac{a\!\cdot\!\sin C}{\sin A}

    . . c \;=\;\frac{2\cdot\dfrac{\sqrt{15}}{4}}{\dfrac{\sqr  t{6}}{4}} \;=\;{\color{blue}\sqrt{10}} \quad\hdots \quad \text{answer A}

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