Question

At 10:00 am, a person observes a hot air balloon climbing vertically in the air from a

point 300 meters away from the launch pad for the balloon. The angle of elevation to
the top of the balloon at this time is 25°. At 10:02am, the person observes that the
angle of elevation to the balloon is now 60°. What is the change in altitude, to the
nearest meter, for the balloon over the 2 minutes between the first and second
observations?

Answer 1

• We are given two observations
• Lets use geometric properties to solve
• 300 meters is the hyp or c
• Theta is 25 degree
• Theta changes to 60 degree
• Hence tan(25)= x1/300
• x1=139.8
• Now let's find the second height
• tan(60) =x2/300
• x2=519.6 ft
• Now x2-x1
• So the difference is 379.8 ft