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

2 Answers
Shwetank Mauria
Jan 14, 2017

#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#

P dilip_k
Jan 14, 2017

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#

Related questions
See all questions in The Law of Cosines
Impact of this question
2323 views around the world
You can reuse this answer
Creative Commons License