Question

a plane flying at a certian altitude is observed from two points that are 3 miles apart. the Angles of elevation made by the two points are 55deg and 72deg​

Answer 1

The altitude of the plane is 8 miles
Step-by-step explanation:
see the attached figure to better understand the problem
we know that
In the right triangle ABC
tan(72°)=h/x
h=xtan(72°) -----> equation A
In the right triangle ABD
tan(55°)=h/(x+3)
h=(x+3)tan(55°) -----> equation B
equate equation A and equation B and solve for x
xtan(72°)=(x+3)tan(55°)
xtan(72°)-xtan(55°)=3tan(55°)
x[tan(72°)-tan(55°)]=3tan(55°)
x=3tan(55°)/[tan(72°)-tan(55°)]
Find the value of h
h=xtan(72°)
h=[3tan(55°)*tan(72°)]/[tan(72°)-tan(55°)]
h=8 miles