Question

Suppose each edge of the cube shown in the figure is x inches long. Find the sine and cosine of the angle formed by diagonals CF and CH.

Answer 1

Answer: The answer is
`(1)/(\sqrt3),~~\sqrt (2)/(3).`

Step-by-step explanation: Given in the question that ABCDEFGH is a cube with edge length 'x' inches. We need to find the sine and cosine of the angle formed by the diagonals CF and CH.

As shown in the attached figure, we cut ΔCFH from the cube, which is a right-angled triangle at ∠CHF=90°. Let, ∠FCH = α.

Now, we need to find the sine and cosine of α.

Since each side of a cube is a square, so CDHG is a square and CH is its diagonal. So,

`CH^2=CD^2+DH^2\\\\\Rightarrow CH^2=x^2+x^2\\\\\Rightarrow CH=\sqrt 2x.`

Again, from right-angled triangle CFH, we have

`CF^2=CH^2+FH^2=(\sqrt2x)^2+x^2=3x^2\\\\\Rightarrow CF=\sqrt3x.`

Therefore,

`\sin \alpha=(FH)/(CF)=(x)/(\sqrt3x)=(1)/(\sqrt3),`

and

`\cos \alpha=(CH)/(CF)=(\sqrt2x)/(\sqrt3x)=\sqrt{(2)/(3)}.`

Thus, the answer is
`(1)/(\sqrt3),~~\sqrt (2)/(3).`