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2 years ago

8

How are the measures of angles, arcs and segments formed by secant lines that intersect within a circle related?

2 answers:

Evgen [1.6K]2 years ago

8 0

The angles and segments formed the same lines

Drupady [299]2 years ago

6 0

Answer:

The answers below ._.

Step-by-step explanation:

A secant is a line that intersects a circle in exactly two points. When a tangent and a secant, two secants, or two tangents intersect outside a circle then the measure of the angle formed is one-half the positive difference of the measures of the intercepted arcs.

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2 years ago

Can you prove it??<br>it's hard,I tried but couldn't solve this.

BARSIC [14]

I'll abbreviate $s=\sin\theta$ and $c=\cos\theta$ , so the identity to prove is

$\dfrac{s+c+1}{s+c-1}-\dfrac{1+s-c}{1-s+c}=2\left(1+\dfrac1s\right)$

On the left side, we can simplify a bit:

$\dfrac{s+c+1}{s+c-1}=\dfrac{s+c-1+2}{s+c-1}=1+\dfrac2{s+c-1}$

$\dfrac{1+s-c}{1-s+c}=-\dfrac{-2+1-s+c}{1-s+c}=-1+\dfrac2{1-s+c}$

Then

$\dfrac{s+c+1}{s+c-1}-\dfrac{1+s-c}{1-s+c}=2\left(1+\dfrac1{s+c-1}-\dfrac1{1-s+c}\right)$

So the establish the original equality, we need to show that

$\dfrac1{s+c-1}-\dfrac1{1-s+c}=\dfrac1s$

Combine the fractions:

$\dfrac{(1-s+c)-(s+c-1)}{(s+c-1)(1-s+c)}=\dfrac{2-2s}{c^2-s^2+2s-1}$

We can rewrite the denominator as

$c^2-s^2+2s-1=c^2+s^2-2s^2+2s-1$

then using the fact that $c^2+s^2=\cos^2\theta+\sin^2\theta=1$ , we get

$1-2s^2+2s-1=2s-2s^2$

so that we have

$\dfrac1{s+c-1}-\dfrac1{1-s+c}=\dfrac{2-2s}{2s-2s^2}=\dfrac1s$

as desired.

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3 years ago