Question-135976

Question Number 135976 by I want to learn more last updated on 17/Mar/21

Answered by mr W last updated on 17/Mar/21

(x/(sin 90°))=(2/(sin ∠ADC)) ...(i) ((sin 30°)/3)=((sin ∠BDC)/y) ...(ii) sin ∠BDC=sin (180°−∠ADC)=sin ∠ADC (i)×(ii): (x/(sin 90°))×((sin 30°)/3)=(2/y) ⇒xy=((2×3×sin 90°)/(sin 30°))=(6/(1/2))=12

$$\frac{{x}}{\mathrm{sin}\:\mathrm{90}°}=\frac{\mathrm{2}}{\mathrm{sin}\:\angle{ADC}}\:\:\:\:…\left({i}\right) \\ $$$$\frac{\mathrm{sin}\:\mathrm{30}°}{\mathrm{3}}=\frac{\mathrm{sin}\:\angle{BDC}}{{y}}\:\:\:…\left({ii}\right) \\ $$$$\mathrm{sin}\:\angle{BDC}=\mathrm{sin}\:\left(\mathrm{180}°−\angle{ADC}\right)=\mathrm{sin}\:\angle{ADC} \\ $$$$\left({i}\right)×\left({ii}\right): \\ $$$$\frac{{x}}{\mathrm{sin}\:\mathrm{90}°}×\frac{\mathrm{sin}\:\mathrm{30}°}{\mathrm{3}}=\frac{\mathrm{2}}{{y}} \\ $$$$\Rightarrow{xy}=\frac{\mathrm{2}×\mathrm{3}×\mathrm{sin}\:\mathrm{90}°}{\mathrm{sin}\:\mathrm{30}°}=\frac{\mathrm{6}}{\frac{\mathrm{1}}{\mathrm{2}}}=\mathrm{12} \\ $$

Commented by I want to learn more last updated on 17/Mar/21

$$\mathrm{Thanks}\:\mathrm{sir}.\:\mathrm{I}\:\mathrm{appreciate}. \\ $$

Commented by I want to learn more last updated on 17/Mar/21

Sir please tag your question on projectile that an object strike the wall and bounced back.

$$\mathrm{Sir}\:\mathrm{please}\:\mathrm{tag}\:\mathrm{your}\:\mathrm{question}\:\mathrm{on}\:\mathrm{projectile}\:\mathrm{that}\:\mathrm{an}\:\mathrm{object}\:\mathrm{strike}\:\mathrm{the} \\ $$$$\mathrm{wall}\:\mathrm{and}\:\mathrm{bounced}\:\mathrm{back}. \\ $$

Commented by mr W last updated on 17/Mar/21