Your three problems here involve right triangles.

For the 1st one,

hypotenuse = unknown

vertical leg = 10

horizontal leg = 100

You are to find the angle opposite the 10-leg.

tan(theta) = 10/100 = 0.1

theta = arctan(0.1)

theta = 5.71 degrees -----answer.

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For the 2nd one,

hypotenuse = unknown

vertical leg = 230 ft/min

horizontal leg = 76 mi/hr

You are to find the angle opposite the 230 ft/min -leg.

First, make sure that the units of measurements must be the same. So convert the 76 mi/hr into ft/min (or the 320 ft/min into mi/hr.)

There are 5280 ft per 1 mile, and 60 min per 1 hour, so,

(230ft/min)*(1mi/5280ft)*(60min/1hr) = (230*1*60)/(5280*1) = 2.613636 mi/hr.

Then, tan(theta) = 2.613636 / 76 = 0.03439

theta = arctan(0.03439)

theta = 1.9696 degrees. --------answer.

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For the 3rd one,

hypotenuse = 400 mi/hr

vertical leg = 10,000 ft/min

horizontal leg = unknown

You are to find the angle opposite the 10,000 ft/min leg.

Same units again,

(10,000ft/min)*(1mi/5280ft)*(60min/1hr) = (10,000*1*60)/(5280*1) = 113.63636 mi/hr.

Then, sin(theta) = 113.63636 / 400 = 0.28409

theta = arcsin(0.03439)

theta = 16.5 degrees. --------answer.