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Kruka [31]
2 years ago
11

Help please!!

Mathematics
2 answers:
Free_Kalibri [48]2 years ago
7 0
Tan²( θ ) - (1 + √3) tan (θ) + √3 = 0

tan²( θ ) - (tan (θ) + √3 tan (θ)) + √3 = 0

tan²( θ ) - tan (θ) - √3 tan (θ) + √3 = 0

tan( θ ) ( tan (θ) - 1) - √3 ( tan (θ) - 1 ) = 0

( tan( θ ) - 1 ) ( tan( θ ) - √3 ) = 0


tan( θ ) - 1 = 0

θ = π/₄

tan( θ ) - √3 = 0


θ = π/₃


so θ = π/₄ and θ = π/₃
irinina [24]2 years ago
7 0

Answer:

\huge \boxed{  \red{  \boxed{\begin{cases} \theta   =   {45}^{ \circ} \\    \theta=      {60}^{ \circ}  \end{cases} }}}

Step-by-step explanation:

<h3>to understand this</h3><h3>you need to know about:</h3>
  • trigonometry
  • PEMDAS
<h3>let's solve:</h3>

distribute tan(θ):

=>tan²(θ)-(tan(θ)+√3tan(θ))+√3=0

remove parentheses:

=>tan²(θ)-tan(θ)-√3tan(θ)+√3=0

so this equation is now in standard form i.e ax²+bx+c=0

we can solve by factoring as we solve quadratic equation

factor out tanθ:

=>tan(θ)(tan(θ)-1)-√3tan(θ)+√3=0

factor out -√3:

=>tan(θ)(tan(θ)-1)-√3(tan(θ)-1)=0

group:

=>(tan(θ)-√3)(tan(θ)-1)=0

separate it as two different equation:

\implies  \begin{cases} \tan( \theta)  - 1 = 0 \\   \tan( \theta)  -  \sqrt{3} = 0  \end{cases}

add 1 and √3 to both sides to first and second equation respectively:

\implies  \begin{cases} \tan( \theta)  - 1 + 1 = 0 + 1 \\   \tan( \theta)  -  \sqrt{3} +  \sqrt{3}  = 0   +  \sqrt{3} \end{cases}

\implies  \begin{cases} \tan( \theta)   =  1 \\   \tan( \theta)  =     \sqrt{3} \end{cases}

\therefore  \begin{cases} \theta   =   {45}^{ \circ} \\    \theta=      {60}^{ \circ}  \end{cases}

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