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Question-217949




Question Number 217949 by hardmath last updated on 23/Mar/25
Commented by hardmath last updated on 23/Mar/25
x^2  + y^2  + z^2  = ?
$$\mathrm{x}^{\mathrm{2}} \:+\:\mathrm{y}^{\mathrm{2}} \:+\:\mathrm{z}^{\mathrm{2}} \:=\:? \\ $$
Answered by A5T last updated on 23/Mar/25
((sin90°)/x)=((sin70°)/1)⇒x=(1/(sin 70°))  Similarly, we get y=(1/(sin 50°)) ; z=(1/(sin10°))   ⇒x^2 +y^2 +z^2 =(1/(sin^2 50°))+(1/(sin^2 10°))+(1/(sin^2 70))=36
$$\frac{\mathrm{sin90}°}{\mathrm{x}}=\frac{\mathrm{sin70}°}{\mathrm{1}}\Rightarrow\mathrm{x}=\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{70}°} \\ $$$$\mathrm{Similarly},\:\mathrm{we}\:\mathrm{get}\:\mathrm{y}=\frac{\mathrm{1}}{\mathrm{sin}\:\mathrm{50}°}\:;\:\mathrm{z}=\frac{\mathrm{1}}{\mathrm{sin10}°}\: \\ $$$$\Rightarrow\mathrm{x}^{\mathrm{2}} +\mathrm{y}^{\mathrm{2}} +\mathrm{z}^{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{sin}^{\mathrm{2}} \mathrm{50}°}+\frac{\mathrm{1}}{\mathrm{sin}^{\mathrm{2}} \mathrm{10}°}+\frac{\mathrm{1}}{\mathrm{sin}^{\mathrm{2}} \mathrm{70}}=\mathrm{36} \\ $$

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