In $triangle XYZ$ if x=1, y=2, $z=sqrt5$ , how do you find the exact value of cosZ?

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In $triangle XYZ$ if x=1, y=2, $z=sqrt5$ , how do you find the exact value of cosZ?

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Answer 1 · 2017-01-14T06:55:17

$cosZ=0$

Explanation:

As $x=1$ , $y=2$ and $z=sqrt5$ ,

it is apparent that the square on the largest side $z=sqrt5$ is $5$ and is equal to sum of the squares on other two sides of the triangle as $5=1^2+2^2$

Hence $z$ is hypotenuse and $m/_Z=pi/2$

and $cosZ=0$

Answer 2 · 2017-01-14T13:08:49

Applying cosine law for triangle we can write

$cosZ=(x^2+y^2-z^2)/(2xy)$

$=>cosZ=(1^2+2^2-(sqrt5)^2)/(2*1*2)=(5-5)/4=0$

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In $triangle XYZ$ if x=1, y=2, $z=sqrt5$ , how do you find the exact value of cosZ?

2 Answers

Explanation:

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Science

Math

Humanities

... and beyond

In triangle XYZtriangle XYZ if x=1, y=2, z=sqrt5z=sqrt5, how do you find the exact value of cosZ?

2 Answers

Explanation:

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Impact of this question

In $triangle XYZ$ if x=1, y=2, $z=sqrt5$ , how do you find the exact value of cosZ?