Two planes leave simultaneously from Chicago's O'Hare Airport, one flying due north and the other due east (see figure). The northbound plane is flying 50 miles per hour faster than the eastbound plane. After 3 hours, the planes are d = 2018 miles apart. Find the speed of each plane. (Round your answers to the nearest whole number.)

a). 170 miles

b). Planes will be 500 miles apart after 1.1 hours.

Step-by-step explanation:

a). Two planes heading towards North and South when passed each other were 80 miles apart.

Distance between these planes after 20 minutes or hours.

From the figure attached,

In right angle triangle DEC,

DE = 80 miles

BE = (Speed × duration) = miles

Similarly, BC = miles

By Pythagoras theorem,

DC² = EC² + DE²

      = (EB + BC)² + DE²

      = (100 + 50)² + (80)²

      = 28900

DC = √28900 = 170 miles

b). Now we have to evaluate the duration after which distance between the planes is 500 miles.

Let after t hours planes will be 500 miles apart.

Then EB = 300t

BC = 150t

Therefore, EC = EB + BC = 450t

It's given that DC = 500 miles

By Pythagoras theorem again,

DC² = EC²+ DE²

(500)²=  (450t)²+ (80)²

250000 = 202500t² + 6400

2500 = 2025t² + 64

2025t² = 2436

t² = 1.20297

t = 1.097 hours ≈ 1.1 hours

Therefore, both the planes will be 500 miles apart after 1.1 hours.

Step-by-step explanation:

we know that

applying the law of sines

°

so

∠j=18°

∠k=29°

∠l=180-(18+29)=133°

the answer part a) is

the measure of angle k is  ≈29°

part b)  the measure of angle k could also be

(180-29)=151°

so

∠j=18°

∠k=151°

∠l=180-(18+151)=11°

the answer part b) is

the measure of angle k could also be 151°

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step-by-step explanation: