Menu Close

Question-139654




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

Leave a Reply

Your email address will not be published. Required fields are marked *