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vitfil [10]
2 years ago
6

CHOOSE WHICH IS CORRECT

Geography
1 answer:
LenaWriter [7]2 years ago
5 0

If a ray is an angle bisector,  then it cuts an angle into two smaller, congruent angles.

For better understanding, let us explain what the answer means

  • This statement 'If a ray cuts an angle into two smaller, congruent angles, then it is an angle" can be rewritten as the answer above but not like the other options because it will end up contradicting itselfs. That is because a ray (one ray) is an angle bisector and if it has to change, it then have to change to two  which is then called an angle trisector ( Two rays that divide an angle into three congruent angles trisect the angle) so it will no longer be one but two and then that we change the statement completely. so we have to go with the second options in the picture shown

From the above, we can therefore say that the answer If a ray is an angle bisector,  then it cuts an angle into two smaller, congruent angles, is correct

Learn more about rays as a bisector from:

brainly.com/question/24485009

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What is the value of X6? <br><br> Show the solution.
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Answer : The value of x_6 is \sqrt{7}.

Explanation :

As we are given 6 right angled triangle in the given figure.

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x_1=\sqrt{(1)^2+(1)^2}

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Using Pythagoras theorem in triangle 2 :

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Now we have to calculate the value of x_3.

Using Pythagoras theorem in triangle 3 :

(Hypotenuse)^2=(Perpendicular)^2+(Base)^2

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(x_3)^2=(1)^2+(\sqrt{3})^2

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Now we have to calculate the value of x_4.

Using Pythagoras theorem in triangle 4 :

(Hypotenuse)^2=(Perpendicular)^2+(Base)^2

(x_4)^2=(1)^2+(X_3)^2

(x_4)^2=(1)^2+(\sqrt{4})^2

x_4=\sqrt{(1)^2+(\sqrt{4})^2}

x_4=\sqrt{5}

Now we have to calculate the value of x_5.

Using Pythagoras theorem in triangle 5 :

(Hypotenuse)^2=(Perpendicular)^2+(Base)^2

(x_5)^2=(1)^2+(X_4)^2

(x_5)^2=(1)^2+(\sqrt{5})^2

x_5=\sqrt{(1)^2+(\sqrt{5})^2}

x_5=\sqrt{6}

Now we have to calculate the value of x_6.

Using Pythagoras theorem in triangle 6 :

(Hypotenuse)^2=(Perpendicular)^2+(Base)^2

(x_6)^2=(1)^2+(X_5)^2

(x_6)^2=(1)^2+(\sqrt{6})^2

x_6=\sqrt{(1)^2+(\sqrt{6})^2}

x_6=\sqrt{7}

Therefore, the value of x_6 is \sqrt{7}.

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