Answer:
To prove:
X+Y.Z=(X+Y).(X+Z)
Taking R.H.S
= (X+Y).(X+Z)
By distributive law
= X.X+X.Z+X.Y+Y.Z --- (1)
From Boolean algebra
X.X = X
X.Y+X.Z = X.(Y+Z)
Using these in (1)
=X+X(Y+Z)+Y.Z
=X(1+(Y+Z)+Y.Z --- (2)
As we know (1+X) = 1
Then (2) becomes
=X.1+Y.Z
=X+Y.Z
Which is equal to R.H.S
Hence proved,
X+Y.Z=(X+Y).(X+Z)
First, let's expand the right side. Now our equation is

Now, we gather all the terms with x in it on the left side, and the numbers without x on the right. We get:

And then we can make this

We divide both sides by 12 to get x on its own, so x = 13/12 or 1.08
Well there has to be a right angle in there somewhere as the pole must be perpendicular to the ground.
Using the sine rule:
sinA/a = sinB/b where a = 10, A = 90, b = 5√3, B = B
Re-arrange for B, B = arcsin (b(sinA)/a)
B = arcsin(5√3(sin90)/10)
B = 60 degrees.
Answer is C
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Step-by-step explanation:
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