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Question Number 139654 by otchereabdullai@gmail.com last updated on 30/Apr/21

Commented by mr W last updated on 30/Apr/21

too less information to find x! you should give e.g. AC or DC.

$${too}\:{less}\:{information}\:{to}\:{find}\:{x}! \\ $$$${you}\:{should}\:{give}\:{e}.{g}.\:{AC}\:{or}\:{DC}. \\ $$

Commented by mr W last updated on 30/Apr/21

Commented by mr W last updated on 30/Apr/21

DC=(h/(tan α)) x+DC=(h/(tan β)) ⇒x=((1/(tan β))−(1/(tan α)))h

$${DC}=\frac{{h}}{\mathrm{tan}\:\alpha} \\ $$$${x}+{DC}=\frac{{h}}{\mathrm{tan}\:\beta} \\ $$$$\Rightarrow{x}=\left(\frac{\mathrm{1}}{\mathrm{tan}\:\beta}−\frac{\mathrm{1}}{\mathrm{tan}\:\alpha}\right){h} \\ $$

Commented by mr W last updated on 30/Apr/21

Commented by mr W last updated on 30/Apr/21

AC=b tan α x+b=((AC)/(tan β))=((tan α)/(tan β))×b ⇒x=(((tan α)/(tan β))−1)b

$${AC}={b}\:\mathrm{tan}\:\alpha \\ $$$${x}+{b}=\frac{{AC}}{\mathrm{tan}\:\beta}=\frac{\mathrm{tan}\:\alpha}{\mathrm{tan}\:\beta}×{b} \\ $$$$\Rightarrow{x}=\left(\frac{\mathrm{tan}\:\alpha}{\mathrm{tan}\:\beta}−\mathrm{1}\right){b} \\ $$

Commented by otchereabdullai@gmail.com last updated on 30/Apr/21

Wow wow wow! nice one prof thanks

$$\mathrm{Wow}\:\mathrm{wow}\:\mathrm{wow}!\:\mathrm{nice}\:\mathrm{one}\:\mathrm{prof}\:\mathrm{thanks} \\ $$

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