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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

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

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

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

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

$$Q135558$$

Commented by otchereabdullai@gmail.com last updated on 17/Mar/21

$$\mathrm{fantastic}!$$

Commented by I want to learn more last updated on 18/Mar/21

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

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