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3 years ago

Given: measure of arc PIV = 7/2 times measure of arc PKV PKV Find: m∠VPJ

Mathematics

1 answer:

Serhud [2]3 years ago

6 0

Answer:

Measure of angle 'VPJ' is 140 degrees.

Step-by-step explanation:

Given:

Measure of arc 'PIV' and measure of arc 'PKV'.

'PIV' = 7/2 times of 'PKV'

Lets say that PKV is 'x'.

⇒ $m(PIV)=\frac{7}{2} \times x$

⇒ $m(PIV)=3.5x$

Note:

A full circle has an arc angle measure of 360.

So,

⇒ $3.5x+x=360$

⇒ $4.5x=360$

⇒ $x=\frac{360}{4.5}$

⇒ $x=80$

The measure of arc 'PKV' = 80 degrees.

We have to find angle 'VPJ' that is having a linear pair with angle 'VPL'.

So before finding we 'VPJ' have to find 'VPL'.

And

According to the theorem:

The angle formed by the tangent and the chord is half the measure of the intercepted arc.

Then.

⇒ $\angle VPL=\frac{m\ arc\ (PKV)}{2}$

⇒ $\angle VPL=\frac{80}{2}$

⇒ $\angle VPL=40$ degrees

⇒ And from linear pair .

⇒ $m\angle VPL +m \angle VPJ =180$

⇒ $40 +m \angle VPJ =180$

⇒ $m \angle VPJ =180-40$

⇒ $m \angle VPJ =140$ degrees.

So measure of angle 'VPJ' is 140 degrees.

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This statement can be proven by contradiction for $n \in \mathbb{N}$ (including the case where $n = 0$ .)

$\text{Let $n \in \mathbb{N}$ be a perfect square}$ .

$\textbf{Case 1.} ~ \text{n = 0}$ :

$\text{$n + 2 = 2$, which isn't a perfect square}$ .

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$\text{Assume that $n$ is a perfect square}$ .

$\text{$\iff$ $\exists$ $a \in \mathbb{N}$ s.t. $a^2 = n$}$ .

$\text{Assume $\textit{by contradiction}$ that $(n + 2)$ is a perfect square}$ .

$\text{$\iff$ $\exists$ $b \in \mathbb{N}$ s.t. $b^2 = n + 2$}$ .

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$\text{$a,\, b \in \mathbb{N} \subset \mathbb{Z}$ $\implies b - a = b + (- a) \in \mathbb{Z}$}$ .

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Step-by-step explanation:

Assume that the natural number $n \in \mathbb{N}$ is a perfect square. Then, (by the definition of perfect squares) there should exist a natural number $a$ ( $a \in \mathbb{N}$ ) such that $a^2 = n$ .

Assume by contradiction that $n + 2$ is indeed a perfect square. Then there should exist another natural number $b \in \mathbb{N}$ such that $b^2 = (n + 2)$ .

Note, that since $(n + 2) > n \ge 0$ , $\sqrt{n + 2} > \sqrt{n}$ . Since $b = \sqrt{n + 2}$ while $a = \sqrt{n}$ , one can conclude that $b > a$ .

Keep in mind that both $a$ and $b$ are natural numbers. The minimum separation between two natural numbers is $1$ . In other words, if $b > a$ , then it must be true that $b \ge a + 1$ .

Take the square of both sides, and the inequality should still be true. (To do so, start by multiplying both sides by $(a + 1)$ and use the fact that $b \ge a + 1$ to make the left-hand side $b^2$ .)

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Expand the right-hand side using the binomial theorem:

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