Part A) Find BC, the distance from Tower 2 to the plane, to the nearest foot.
in the triangle ACD
sin16=CD/(7600+BD)--------> CD=sin16*(7600+BD)---------> equation 1
in the triangle BCD
sin24=CD/BD-----------> CD=sin24*BD---------------> equation 2
equation 1=equation 2
sin16*(7600+BD)=sin24*BD-----> sin16*7600+sin16*BD=sin24*BD
sin24*BD-sin16*BD=sin16*7600----> BD=[sin16*7600]/[sin24-sin16]
BD=15979 ft
in the triangle BCD
cos24=BD/BC---------> BC=BD/cos24-------> 15979/cos24-------> 17491
BC=17491 ft
the answer part 1) BC is 17491 ft
Part 2) Find CD, the height of the plane from the ground, to the nearest foot.
CD=sin24*BD ( remember equation 2)
BD=15979 ft
CD=sin24*15979 -----------> CD=6499 ft
the answer part 2) CD is 6499 ft
Answer:
A. 10 B. 15
Step-by-step explanation:
A. 25/1 * 2/5= 50/5
reduced it is 10
B. 25-10=15
The average of the given base lengths is the length of the midsegment:
... (72 + 104)/2 = 176/2 = 88
