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[Solved] Triangle area trigonometry

 
Trigonometry 1 & 2
Last Post by Staff 4 years ago
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June 15, 2021 2:20 pm
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Staff 
(@armor)
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June 15, 2021 2:20 pm
Topic starter
Staff 
(@armor)
Admin
  Honorable Member
317 Posts
156 153 0

Since  \small sinA+cosA=\sqrt{2}cos(A-45^{\circ})=\frac{\sqrt{2}}{2}\Rightarrow cos(A-45^{\circ})=\frac{1}{2}.

Additionally, \small 0< A< 180^{\circ},  then  \small A-45^{\circ}=60^{\circ}\Rightarrow A=105^{\circ}

\small \Rightarrow tanA=tan(45^{\circ}+60^{\circ})=\frac{1+\sqrt{3}}{1-\sqrt{3}}=-2-\sqrt{3}.

Since   \small sinA=sin(45^{\circ}+60^{\circ})=sin45^{\circ}cos60^{\circ}+cos45^{\circ}sin60^{\circ}=\frac{\sqrt{2}+\sqrt{6}}{4},

we have    \small S(\angle ABC)=\frac{1}{2}.AC.AB.sinA=\frac{1}{2}.2.3.\frac{\sqrt{2}+\sqrt{6}}{4}=\frac{3}{4}(\sqrt{2}+\sqrt{6}).

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