The angles of elevation of an airplane between two observation points, a and b, are 30 ° and 45 ° respectively. if the vertical height of the plane at the moment is 5 km, find the distance between a and b to the nearest tenth of a km.

xA = 10 km

xB = 7.071km

Step-by-step explanation:

A/Q

p=5km

For point A

∅ = 30°

sin ∅ = p/h

sin 30° = 5/h

1/2 = 5/h

h = 5×2

h=10km

For point B

sin ∅ = p/h

sin 45° = 5/x

1/√2 = 5/x

x = 5 × √2

x = 7.071 km

.: Points A and B are 10 km and 7.071 km away from the airplane.

area of part 1:

a1=base*height/2=(8+2)*(8-2-2-1)/2=10*3/2=15 ft^2

area of part 2:

a2=length*width=(8+2)*1=10 ft2

area of part 3:

a3=length*width=8*2=16 ft2

total shaded area : a1+a2+a3=41 ft2

total area of the reactangle : a=16*8=128 ft2

total area of the nonshaded region : 128-41=87 ft2

what is the graph

step-by-step explanation: