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2 years ago

59. In the given figure, O is the centre of circle, XY || BC, what is the value of x?

Mathematics

1 answer:

likoan [24]2 years ago

6 0

The value of x is 25 degrees if XY is parallel to the BC option (b) 25 degrees is correct.

<h3>What is a circle?</h3>

It is described as a set of points, where each point is at the same distance from a fixed point (called the center of a circle)

It is given that:

A circle is given:

XY ║ BC

x = angle XYB

When two lines or rays converge at the same point, the measurement between them is called an "Angle."

Angle BAX = Angle XYB

In triangle AXY, Angle XAY = 90 degrees

Angle BAX = 90 - Angle BAY

x = Angle BAX = 90 - 65

x = 25 degrees

Thus, the value of x is 25 degrees if XY is parallel to the BC option (b) 25 degrees is correct.

Learn more about circle here:

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Answer:

See proof below

Step-by-step explanation:

We will use properties of inequalities during the proof.

Let $y\in (x-\epsilon,x+\epsilon)$ . then we have that $x-\epsilon$ . Hence, it makes sense to define the positive number delta as $\delta=\min\{x+\epsilon-y,y-(x-\epsilon)\}$ (the inequality guarantees that these numbers are positive).

Intuitively, delta is the shortest distance from y to the endpoints of the interval. Now, we claim that $(y-\delta,y+\delta)\subseteq (x-\epsilon,x+\epsilon)$ , and if we prove this, we are done. To prove it, let $z\in (y-\delta,y+\delta)$ , then $y-\delta$ . First, $\delta \leq y-(x-\epsilon)$ then $-\delta \geq -y+x-\epsilon$ hence $z>y-\delta \geq x-\epsilon$

On the other hand, $\delta \leq x+\epsilon-y$ then $z$ hence $z$ . Combining the inequalities, we have that $x-\epsilon$ , therefore $(y-\delta,y+\delta)\subseteq (x-\epsilon,x+\epsilon)$ as required.

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4 years ago

In a right triangle ABC, CD is an altitude, such that AD=BC. Find AC, if AB=3 cm, and CD= 2 cm.

konstantin123 [22]

Consider right triangle ΔABC with legs AC and BC and hypotenuse AB. Draw the altitude CD.

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According to this theorem,

$BC^2=BD\cdot AB.$