Solution :
We observe that :
But BA is the perpendicular.
From the center B and WX is a chord.
Therefore, TW = TX (perpendicular from the centre of a circle to a chord bisects it)
Consider Δ BTX,
∠BTX = 90° (BA ⊥ WX)
BT = XT (Δ BTX is isosceles)
Since the angles opposite to equal sides are equal of a triangle arc are equal.
∠BTX = ∠BXT
But in the triangle,
∠TBX + ∠TXB + ∠BTX = 180°
∠TBX + ∠TBX + 90° = 180°
2 ∠TBX = 90°
∠TBX = 45°
From trigonometry, we get
...............(1)
WX = 10
i.e., TX + TW = 10
But TX = TW
2 TX = 10
Tx = 5
BX = radius of circle.
∴ 
= 7
Therefore, the radius of the circle is 7 units.
Answer:
f(t) = 15.06 m
Step-by-step explanation:
The relationship between the height of the rocket and the time after launch, t. In seconds is given by :
We need to find the maximum height the rocket will reach the ground.
When it reaches the ground,
f(t) = 0
So,
Maximum height,
f'(t) = 0
Put t = 1.63 in equation (1).
So, the maximum height of the rocket is 15.06 m.
1.0 and 1/1
100% is the same as 1.
Answer: V = 366917713920 cm³
Based on definition of <em>unit</em> circle and <em>angle</em> properties on <em>Cartesian</em> plane, we find the following results: A: π/5, B: 3π/7, C: 4π/3.
<h3>How to determine angles in a unit circle</h3>
<em>Units</em> circles are commonly used to understand angles and <em>trigonometric</em> functions in a simple way. Vectors in a <em>unit</em> circle are <em>ordered</em> pairs of <em>polar</em> form: (x, y) = (cos θ, sin θ).
To solve this problem, we must take these tips into account:
- Angles in the <em>first</em> quadrant are within 0 < θ < π/2.
- Angles in the <em>second</em> quadrant are within π/2 < θ < π.
- Angles in the <em>third</em> quadrant are within π < θ < 3π/2.
- Angles in the <em>fourth</em> quadrant are within 3π/2 < θ < 2π.
Then, we have the following results: A: π/5, B: 3π/7, C: 4π/3.
To learn more on angles: brainly.com/question/13954458
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