John-stands-at-10m-from-a-mango-tree-while-Philip-stands-between-John-and-the-Philip-The-angle-of-elevation-from-John-and-Philip-is-55-and-70-respectively-If-John-is-1-7m-tall-and-Philip-is-1-5m-

Question Number 190449 by MathsFan last updated on 03/Apr/23

$$ \\ $$John stands at 10m from a mango tree while Philip stands between John and the Philip. The angle of elevation from John and Philip is 55º and 70º respectively. If John is 1.7m tall and Philip is 1.5m tall, find the minimum length of stick they will each need to touch the mango from their position.
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Answered by a.lgnaoui last updated on 14/Apr/23

Donnes: AC=1,7m EF=1,5m CD=10 ∡BAH=55° ∡DED=70° BD=BH+HD=AC+HB tan 55=((HB)/(10))⇒HB=10tan 55 ⇒BD=10tan 55+1,7=15,9m 2•length of stick distant from position C for John to touch tree is:(BC) BC=(√(CD^2 +BD^2 )) BC=(√(10^2 +15,9^2 )) =18,78m 2•the length of stik from position of Philip to touch tree is (BE) BE^2 =DE^2 +BD^2 ; ((DE)/(BE))=cos 70⇒DE=BEcos 70 BE^2 =(BEcos 70)^2 +BD^2 BE^2 sin^2 70 =BD^2 BE=((BD)/(sin 70)) BE=19,98m (suite : justification donnes λ) tan α=((BD)/(CD))=((15,9)/(10))=1,6 α=58 β=110⇒∡CBE=12 ⇒λ=∡BED=58+12=70

$${Donnes}: \\ $$$${AC}=\mathrm{1},\mathrm{7}{m}\:\:\:\:\:{EF}=\mathrm{1},\mathrm{5}{m}\:\:{CD}=\mathrm{10}\: \\ $$$$\measuredangle{BAH}=\mathrm{55}°\:\:\:\measuredangle{DED}=\mathrm{70}° \\ $$$${BD}={BH}+{HD}={AC}+{HB} \\ $$$$\mathrm{tan}\:\:\mathrm{55}=\frac{{HB}}{\mathrm{10}}\Rightarrow{HB}=\mathrm{10tan}\:\mathrm{55} \\ $$$$\Rightarrow{BD}=\mathrm{10tan}\:\mathrm{55}+\mathrm{1},\mathrm{7}=\mathrm{15},\mathrm{9}{m} \\ $$$$\:\mathrm{2}\bullet{length}\:{of}\:{stick}\:{distant}\:{from} \\ $$$$\:{position}\:{C}\:{for}\:{John}\:{to}\:{touch}\:{tree}\:{is}:\left({BC}\right) \\ $$$$\:{BC}=\sqrt{{CD}^{\mathrm{2}} +{BD}^{\mathrm{2}} } \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{BC}}=\sqrt{\mathrm{10}^{\mathrm{2}} +\mathrm{15},\mathrm{9}^{\mathrm{2}} \:}\:=\mathrm{18},\mathrm{78}{m} \\ $$$$ \\ $$$$\:\mathrm{2}\bullet{the}\:{length}\:{of}\:{stik}\:{from}\:{position} \\ $$$$\:\:{of}\:{Philip}\:\:{to}\:{touch}\:\:{tree}\:{is}\:\left({BE}\right) \\ $$$$\:\:{BE}^{\mathrm{2}} ={DE}^{\mathrm{2}} +{BD}^{\mathrm{2}} \:\:;\:\frac{{DE}}{{BE}}=\mathrm{cos}\:\mathrm{70}\Rightarrow{DE}={BE}\mathrm{cos}\:\mathrm{70} \\ $$$$\:{BE}^{\mathrm{2}} =\left({BE}\mathrm{cos}\:\mathrm{70}\right)^{\mathrm{2}} +{BD}^{\mathrm{2}} \\ $$$$\:{BE}^{\mathrm{2}} \mathrm{sin}^{\mathrm{2}} \mathrm{70}\:={BD}^{\mathrm{2}} \\ $$$$\:\:{BE}=\frac{{BD}}{\mathrm{sin}\:\mathrm{70}}\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{BE}}=\mathrm{19},\mathrm{98}{m} \\ $$$$\left({suite}\::\:{justification}\:{donnes}\:\lambda\right)\: \\ $$$$\mathrm{tan}\:\alpha=\frac{{BD}}{{CD}}=\frac{\mathrm{15},\mathrm{9}}{\mathrm{10}}=\mathrm{1},\mathrm{6} \\ $$$$\alpha=\mathrm{58}\:\:\:\:\beta=\mathrm{110}\Rightarrow\measuredangle{CBE}=\mathrm{12} \\ $$$$\Rightarrow\lambda=\measuredangle{BED}=\mathrm{58}+\mathrm{12}=\mathrm{70} \\ $$

Commented by a.lgnaoui last updated on 04/Apr/23

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