Answer:
79.7°
Step-by-step explanation:
We resolve the speed of the plane into horizontal and vertical components respectively as 300cos75° and 300sin75° respectively. Since the wind blows due west at a speed of 25 miles per hour, its direction is horizontal and is given by 25cos180° = -25 mph. We now add both horizontal components to get the resultant horizontal component of the airplane's speed.
So 300cos75° mph + (-25 mph) = 77.646 - 25 = 52.646 mph.
The vertical component of its speed is 300sin75° since that's the only horizontal motion of the airplane. So the resultant vertical component of the airplane's speed is 300sin75° = 289.778 mph
The direction of the plane, Ф = tan⁻¹(vertical component of speed/horizontal component of speed)
Ф = tan⁻¹(289.778 mph/52.646 mph)
Ф = tan⁻¹(5.5043)
Ф = 79.7°
Answer:
no then its not a right triangle
Step-by-step explanation:
Write out the sum formula for sin
<span>sin(x + y) = sinxcos + sinycosx </span>
<span>Then expand sin(a + b) + sin(a - b) </span>
<span>sinacosb + sinbcosa + sinacosb - sinbcosa </span>
<span>The 2nd and 4th terms cancel and you get </span>
<span>2sinacosb</span>
Answer:
No, similar triangles cannot contain a pair of parallel lines. Yes, because || . Yes, because ∠QUR ≅ ∠TUS (vertical angles) and ∠R ≅ ∠S (alternate interior angles).
Step-by-step explanation:
Uhhh 0.3 or 1.2 lol hope this helps
