Question

An airplane takes off at a speed S of 235235 mph at an angle of 1616degrees° with the horizontal. Resolve the vector S into components.

Answer 1

Answer on Question #70871 – Math – Trigonometry

Question

An airplane takes off at a speed S of 235235 mph at an angle of 1616 degrees° with the horizontal. Resolve the vector S into components.

Solution

One turnover is 360 degrees, so $1616{}^{\circ} = 4 \times 360{}^{\circ} + 176{}^{\circ}$; $|S| = 235235$.

Let CB be perpendicular to AB and CD be perpendicular to DA.

Then AD will be S (x), a horizontal projection of S.

AB will be S (y), a vertical projection of S.

$$\cos(\angle CAD) = AD / AC.$$

So $S(x) = AD = AC \times \cos(\angle CAD) = AC \times \cos(4{}^{\circ}) = 235235 \times 0.9975640 = 234661.979$ (mph);

$$\cos(\angle CAB) = AB / AC.$$

So $AB = AC \times \cos(\angle CAB) = AC \times \cos(86{}^{\circ}) = 235235 \times 0.0697 = 16409.16$ (mph)

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