(From the top of a tower of height 50 m, the angles of depression of t

1.	(From the top of a tower of height 50 m, the angles of depression of the top and bottomof a pole are 30° and 45° respectively. Find the height of the pole. (√3=1.73)
Answer» LetABbethetowerandCDbethepole. Now,AB=50m. Lettheheightofthepole=hmetre. DrawCL⊥AB. Now,LB=hmetresandAL=(50−h)mts LetBD=xmts Now,LC=BD=xmts. In∆ABD, tan45°=ABBD⇒1=50x⇒x=50 In∆ALC,tan30°=ALLC⇒1√3=50−h/x⇒50=503√−3√h⇒3√h=50(3√−1)⇒h=50(3√−1)/√3=50(√3−1)/√3×√3/root 3 =150−50√3/3=50(3−√3)/3=50[3−1.732]/3=50×1.2683=21.13 m

(From the top of a tower of height 50 m, the angles of depression of the top and bottomof a pole are 30° and 45° respectively. Find the height of the pole. (√3=1.73)

Answer»

LetABbethetowerandCDbethepole.

Now,AB=50m.

Lettheheightofthepole=hmetre.

DrawCL⊥AB.

Now,LB=hmetresandAL=(50−h)mts

LetBD=xmts

Now,LC=BD=xmts.

In∆ABD,

tan45°=ABBD⇒1=50x⇒x=50

In∆ALC,tan30°=ALLC⇒1√3=50−h/x⇒50=503√−3√h⇒3√h=50(3√−1)⇒h=50(3√−1)/√3=50(√3−1)/√3×√3/root 3

=150−50√3/3=50(3−√3)/3=50[3−1.732]/3=50×1.2683=21.13 m