A plane decides to travel along AB towards the North with a velocity of V.
1 answer:
Step-by-step explanation:
The wind has a speed of w and a direction α with the vertical. The x component of that speed is w sin α. The y component is -w cos α.
In order to stay on the north trajectory AB, the plane must have a horizontal speed of -w sin α. The plane's speed is v, so using Pythagorean theorem, the y component of the plane's speed is:
v² = (-w sin α)² + vᵧ²
v² = w² sin²α + vᵧ²
vᵧ = √(v² − w² sin²α)
The total vertical speed is therefore √(v² − w² sin²α) − w cos α.
If a is the length of AB, then the time is:
t = a / [√(v² − w² sin²α) − w cos α]
To rationalize the denominator, we multiply by the conjugate.
t = a / [√(v² − w² sin²α) − w cos α] × [√(v² − w² sin²α) + w cos α] / [√(v² − w² sin²α) + w cos α]
t = a [√(v² − w² sin²α) + w cos α] / (v² − w² sin²α − w² cos²α)
t = a [√(v² − w² sin²α) + w cos α] / (v² − w²)
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3. PB is Common, so by the side-side-side congruence postulate, triangles ABP and CBP are congruent.
4. So measure of m(BPA)=x/2 and m(ABP)=73/2.
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2. AB=BC, because tangents drawn to a circle from the same point are equal.
3. PB is Common, so by the side-side-side congruence postulate, triangles ABP and CBP are congruent.
4. So measure of m(BPA)=x/2 and m(ABP)=73/2.
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