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Question Number 144268 by aliibrahim1 last updated on 23/Jun/21

Answered by imjagoll last updated on 24/Jun/21

consider ∡CDA = 90°−α and ∡CAD=90°−α so x = CD=3 cm

$$\mathrm{consider}\:\measuredangle\mathrm{CDA}\:=\:\mathrm{90}°−\alpha\:\mathrm{and}\: \\ $$$$\measuredangle\mathrm{CAD}=\mathrm{90}°−\alpha\:\mathrm{so}\:\mathrm{x}\:=\:\mathrm{CD}=\mathrm{3}\:\mathrm{cm} \\ $$

Commented by som(math1967) last updated on 24/Jun/21

Is ∡BAC=90° ?

$${Is}\:\measuredangle{BAC}=\mathrm{90}°\:?\: \\ $$

Answered by mr W last updated on 24/Jun/21

4^2 =5^2 +x^2 −10x cos 2α ⇒cos 2α=((x^2 +9)/(10x)) AD^2 =3^2 +x^2 −6x cos 2α AD^2 =3^2 +x^2 −6x×((x^2 +9)/(10x)) ⇒AD^2 =((2x^2 +18)/5) 2^2 =AD^2 +4^2 −8ADcos α ((2x^2 +18)/5)+12=8ADcos α ((x^2 +39)/5)=4ADcos α (((x^2 +39)/5))^2 =8×((2x^2 +18)/5)×(cos 2α+1) (x^2 +39)^2 x=8(x^2 +9)(x^2 +10x+9) ⇒x=1,3,9 only x=3 is valid.

$$\mathrm{4}^{\mathrm{2}} =\mathrm{5}^{\mathrm{2}} +{x}^{\mathrm{2}} −\mathrm{10}{x}\:\mathrm{cos}\:\mathrm{2}\alpha \\ $$$$\Rightarrow\mathrm{cos}\:\mathrm{2}\alpha=\frac{{x}^{\mathrm{2}} +\mathrm{9}}{\mathrm{10}{x}} \\ $$$${AD}^{\mathrm{2}} =\mathrm{3}^{\mathrm{2}} +{x}^{\mathrm{2}} −\mathrm{6}{x}\:\mathrm{cos}\:\mathrm{2}\alpha \\ $$$${AD}^{\mathrm{2}} =\mathrm{3}^{\mathrm{2}} +{x}^{\mathrm{2}} −\mathrm{6}{x}×\frac{{x}^{\mathrm{2}} +\mathrm{9}}{\mathrm{10}{x}} \\ $$$$\Rightarrow{AD}^{\mathrm{2}} =\frac{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{18}}{\mathrm{5}} \\ $$$$\mathrm{2}^{\mathrm{2}} ={AD}^{\mathrm{2}} +\mathrm{4}^{\mathrm{2}} −\mathrm{8}{AD}\mathrm{cos}\:\alpha \\ $$$$\frac{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{18}}{\mathrm{5}}+\mathrm{12}=\mathrm{8}{AD}\mathrm{cos}\:\alpha \\ $$$$\frac{{x}^{\mathrm{2}} +\mathrm{39}}{\mathrm{5}}=\mathrm{4}{AD}\mathrm{cos}\:\alpha \\ $$$$\left(\frac{{x}^{\mathrm{2}} +\mathrm{39}}{\mathrm{5}}\right)^{\mathrm{2}} =\mathrm{8}×\frac{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{18}}{\mathrm{5}}×\left(\mathrm{cos}\:\mathrm{2}\alpha+\mathrm{1}\right) \\ $$$$\left({x}^{\mathrm{2}} +\mathrm{39}\right)^{\mathrm{2}} {x}=\mathrm{8}\left({x}^{\mathrm{2}} +\mathrm{9}\right)\left({x}^{\mathrm{2}} +\mathrm{10}{x}+\mathrm{9}\right) \\ $$$$\Rightarrow{x}=\mathrm{1},\mathrm{3},\mathrm{9} \\ $$$${only}\:{x}=\mathrm{3}\:{is}\:{valid}. \\ $$

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