SOLUTION: Please let me know if I'm on the right track with this. A plane is observed approaching your home and you assume the its speed is 550 km/hr. The angle of elevation of the plane i

Question 20855: Please let me know if I'm on the right track with this.
A plane is observed approaching your home and you assume the its speed is 550 km/hr. The angle of elevation of the plane is 16 degrees at one time and 57 degrees one minute later. Approximate the altitude of the plane.
Let r be the approximate altitude of the plane.
41 degrees = 9.17 km
360 degrees = 80.52 km
80.52=2pier
r=80.52/2pie
r=12.82 km
Thank you!
Answer by venugopalramana(3286) (Show Source): You can put this solution on YOUR website!
SEE MY COMMENTS IN CAPITAL LETTERS

A plane is observed approaching your home and you assume the its speed is 550 km/hr. ......OK.....
The angle of elevation of the plane is 16 degrees at one time......OK
and 57 degrees .....OK
one minute later....OK
Approximate the altitude of the plane....OK
Let r be the approximate altitude of the plane.....OK
I AM NOT ABLE TO FOLLOW THE LATER PART AS YOU HAVE NOT GIVEN A SKETCH..BUT IT LOOKS AS IF YOU ASSUMED A CIRCULAR FIGURE OR SO....OK ..LET ME TRY...LET YOUR POSITION ON GROUND BE A.LET THE PLANE BE AT P FIRST AND Q LATER.SO PQ = DISTANCE TRAVELLED BY PLANE IN 1 MINUTE=550/60=55/6 KM.FROM P AND Q DROP PERPENDICULARS PB AND QC TO GROUND LEVEL WHERE YOU ARE.SO PB=QC=ALTITUDE OF PLANE = R ASSUMED.
APB IS A RIGHT ANGLED TRIANGLE WITH ANGLE PAB=16...HENCE TAN 16 =R/AB....
AB=R/TAN(16)
SIMILARLY AC=R/TAN(57)
BC=AB-AC=R/TAN(16)-R/(TAN(57)=R(1/TAN16-1/TAN57)..BUT...BC=PQ=55/6..SO WE HAVE
55/6=R(1/TAN16-1/TAN57)...FROM TABLES YOU CAN FIND TAN 16 AND TAN 57 AND USING THOSE VALUES YOU CAN FIND R FROM THE ABOVE EQN.
41 degrees = 9.17 km
360 degrees = 80.52 km
80.52=2pier
r=80.52/2pie
r=12.82 km
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