50 POINTS (SUPER IMPORTANT PLEASE HELP)
1 answer:
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Given :
- ZC = 90°
- CD is the altitude to AB.
A = 65°.
To find :
- the angles in △CBD and △CAD if m∠A = 65°
Solution :
In Right angle △ABC,
we have,
=> ACB = 90°
=> CAB = 65°.
So,
=> ACB +
CAB+
ZCBA = 180° (By angle sum Property.)
=> 90° + 65° + CBA = 180°
=> 155° +CBA = 180°
=> CBA = 180° - 155°
=> CBA = 25°.
In △CDB,
=> CD is the altitude to AB.
So,
=> CDB = 90°
=> CBD =
CBA = 25°.
So,
=> CBD +
DCB = 180° (Angle sum Property.)
=> 90° +25° + DCB = 180°
=> 115° + DCB = 180°
=> DCB = 180° - 115°
=> DCB = 65°.
Now, in △ADC,
=> CD is the altitude to AB.
So,
=> ADC = 90°
=> CAD =
CAB = 65°.
So,
=> ADC +
CAD +
DCA = 180° (Angle sum Property.)
=> 90° + 65° + DCA = 180°
=> 155° +DCA = 180°
=> DCA = 180° - 155°
=> DCA = 25°
Hence, we get,
Answer:
(-3, 4) is a solution
Step-by-step explanation:
The point (-3, 4) is inside the shaded area of the graph, so is a solution.
You can check in the inequality
y > -2x -3
4 > -2(-3) -3 . . . . substitute for x and y
4 > 3 . . . . . . . true; the given point is a solution
Answer:
The drift angle is approximately 7.65° towards the East from the plane's heading
Step-by-step explanation:
The speed of the plane = 350 mph
The direction in which the plane flies N 40° E = 50° counterclockwise from the eastern direction
The speed of the wind = 40 mph
The direction of the wind = S 70° E = 20° clockwise from the eastern direction
The component velocities of the plane are;
= (350 × cos 50)·i + (350 × sin 50)·j
= (40 + cos 20)·i - (40 × sin 40)·j
The resultant speed of the plane = +
= 265.915·i +242.404·j
The direction the plane is heading = tan⁻¹(242.404/265.915) ≈ 42.35°
Therefore, the drift angle = Actual Angle - Direction of the plane = 50 - 42.35 ≈ 7.65° towards the East