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Question Number 147768 by aliibrahim1 last updated on 23/Jul/21

Answered by liberty last updated on 23/Jul/21

Let BC = x ⇒tan 44°=((CD)/x) ⇒CD=x tan 44° Let AC =y⇒tan 48°=((CD)/y) ⇒CD=y tan 48° where y=(√(30^2 −x^2 )) =(√(900−x^2 )) (•)CD=CD ⇒x tan 44°=(√(900−x^2 )) tan 48° ⇒x^2 tan^2 44°=(900−x^2 )tan^2 48° ⇒x^2 (tan^2 44°+tan^2 48°)=900×tan^2 48° ⇒x = (√((900×tan^2 48°)/(tan^2 44°+tan^2 48°))) ≈22.64 ∴ CD=22.64×tan 44°≈ 21.87

$${Let}\:{BC}\:=\:{x}\:\Rightarrow\mathrm{tan}\:\mathrm{44}°=\frac{{CD}}{{x}} \\ $$$$\Rightarrow{CD}={x}\:\mathrm{tan}\:\mathrm{44}° \\ $$$$\mathrm{L}{et}\:{AC}\:={y}\Rightarrow\mathrm{tan}\:\mathrm{48}°=\frac{{CD}}{{y}} \\ $$$$\Rightarrow{CD}={y}\:\mathrm{tan}\:\mathrm{48}° \\ $$$${where}\:{y}=\sqrt{\mathrm{30}^{\mathrm{2}} −{x}^{\mathrm{2}} }\:=\sqrt{\mathrm{900}−{x}^{\mathrm{2}} } \\ $$$$\left(\bullet\right){CD}={CD} \\ $$$$\Rightarrow{x}\:\mathrm{tan}\:\mathrm{44}°=\sqrt{\mathrm{900}−{x}^{\mathrm{2}} }\:\mathrm{tan}\:\mathrm{48}° \\ $$$$\Rightarrow{x}^{\mathrm{2}} \:\mathrm{tan}\:^{\mathrm{2}} \mathrm{44}°=\left(\mathrm{900}−{x}^{\mathrm{2}} \right)\mathrm{tan}\:^{\mathrm{2}} \mathrm{48}° \\ $$$$\Rightarrow{x}^{\mathrm{2}} \left(\mathrm{tan}\:^{\mathrm{2}} \mathrm{44}°+\mathrm{tan}^{\mathrm{2}} \:\mathrm{48}°\right)=\mathrm{900}×\mathrm{tan}\:^{\mathrm{2}} \mathrm{48}° \\ $$$$\Rightarrow{x}\:=\:\sqrt{\frac{\mathrm{900}×\mathrm{tan}^{\mathrm{2}} \:\mathrm{48}°}{\mathrm{tan}\:^{\mathrm{2}} \mathrm{44}°+\mathrm{tan}\:^{\mathrm{2}} \mathrm{48}°}}\:\approx\mathrm{22}.\mathrm{64} \\ $$$$\therefore\:{CD}=\mathrm{22}.\mathrm{64}×\mathrm{tan}\:\mathrm{44}°\approx\:\mathrm{21}.\mathrm{87} \\ $$

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