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Question-126010




Question Number 126010 by bramlexs22 last updated on 16/Dec/20
Answered by liberty last updated on 16/Dec/20
The area of a tringle is A=(1/2)(8)(6)sin θ  A = 24 sin θ  (dA/dt) = (24 cos θ) (dθ/dt) ; where (dθ/dt) = 0.12 rad/sec  ⇔ (dA/dt) = 24×0.12×cos (π/6)= 2.49 m^2 /sec
$${The}\:{area}\:{of}\:{a}\:{tringle}\:{is}\:{A}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{8}\right)\left(\mathrm{6}\right)\mathrm{sin}\:\theta \\ $$$${A}\:=\:\mathrm{24}\:\mathrm{sin}\:\theta \\ $$$$\frac{{dA}}{{dt}}\:=\:\left(\mathrm{24}\:\mathrm{cos}\:\theta\right)\:\frac{{d}\theta}{{dt}}\:;\:{where}\:\frac{{d}\theta}{{dt}}\:=\:\mathrm{0}.\mathrm{12}\:{rad}/{sec} \\ $$$$\Leftrightarrow\:\frac{{dA}}{{dt}}\:=\:\mathrm{24}×\mathrm{0}.\mathrm{12}×\mathrm{cos}\:\frac{\pi}{\mathrm{6}}=\:\mathrm{2}.\mathrm{49}\:{m}^{\mathrm{2}} /{sec}\: \\ $$$$ \\ $$

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