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Question Number 170624 by cortano1 last updated on 27/May/22

Commented by som(math1967) last updated on 28/May/22

Commented by som(math1967) last updated on 28/May/22

Commented by som(math1967) last updated on 28/May/22

$$Ithink\mathrm{\angle }ABCshouldbe\theta$$

Commented by cortano1 last updated on 28/May/22

$$\mathrm{\angle }ABD=\theta$$

Commented by som(math1967) last updated on 28/May/22

$$if\mathrm{\angle }ABC=\theta thenIcansolvetheproblem$$

Commented by cortano1 last updated on 28/May/22

$$oksir.let\mathrm{\angle }ABC=\theta$$

Commented by som(math1967) last updated on 28/May/22

$$DE=FC=ksin\varphi$$$$BE=kcos\varphi$$$$BC=hcot\theta$$$$\therefore EC=DF=hcot\theta -kcos\varphi$$$$AF=h-FC=h-ksin\varphi$$$$\frac{DF}{AF}=cot\alpha$$$$\frac{hcot\theta -kcos\varphi }{h-ksin\varphi }=cot\alpha$$$$hcot\theta -kcos\varphi =hcot\alpha -kcot\alpha sin\varphi$$$$h(cot\theta -cot\alpha )=k(cos\varphi -sin\varphi cot\alpha )$$$$\mathit{h}=\frac{\mathit{k}(\mathit{c}\mathit{o}\mathit{s}\mathit{\varphi }-\mathit{s}\mathit{i}\mathit{n}\mathit{\varphi }\mathit{c}\mathit{o}\mathit{t}\mathit{\alpha })}{(\mathit{c}\mathit{o}\mathit{t}\mathit{\theta }-\mathit{c}\mathit{o}\mathit{t}\mathit{\alpha })}$$$$$$

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