Question 25074
A window in a bulinging is 18m above the ground. From this window, the angle of elevation to the top of the building across the street is 40° and the angle of depression to the bottom of the building acress the street is 42°. How tall is the building across the street? Include a labelled diagram with you solution
I AM GIVING A GRAPH BELOW FOR YOUR GUIDANCE.YOU CAN COMPLETE THE DRAWING AND LABEL THE DIAGRAM ON THE BASIS GIVEN BELOW.
LET O BE THE ORIGIN;LET THE Y AXIS BE THE BUILDING , THE WINDOW AT 18M.ABOVE GROUND BE ON THE Y AXIS AT W ; AND X AXIS THE GROUND.
HENCE OW = 18 M.
 THE LINE OF ELEVATION OF TOP OF THE BUILDIND ACROSS THE STREET FROM WINDOW ON Y AXIS IS SHOWN BY WT;AND THE LINE OF DEPRESSION OF THE BUILDING FROM THIS WINDOW IS SHOWN BY WB.JOIN BT WHICH WILL BE PERPENDICULAR TO THE GROUND OR X AXIS.DRAW A HORIZONTAL W H FROM WIDOW W TO BUILDING LINE BT.
{{{ graph( 600, 600, 0, 10, 0, 10, x+4,4-x,4) }}}
WE HAVE 2 RIGHT ANGLE TRIANGLES WTH AND WHB.IN TRIANGLE WTH
ANGLE TWH = 40 DEGREES THE ANGLE OF ELEVATION
HENCE TAN 40 = TH/WH  OR TH =WH * TAN 40.............I
IN TRIANGLE WHB 
ANGLE HWB = 42 DEGREES THE ANGLE OF DEPRESSION.
HENCE TAN 42 = HB/WH  OR  WH=HB/TAN 42
BUT HB = OW =18 M ..SO..WH=18 /TAN 42.................II
SUBSTITUTING IN EQN.I WE GET 
TH = 18*TAN 40 / TAN 42
SO HEIGHT OF BUILDING = TH +HB =18*TAN 40 / TAN 42 +18
=18*{(TAN 40 / TAN 42 ) + 1 }..YOU CAN SUBSTITUTE THE VALUES FOR TAN 40  AN TAN 42 AND GET