Given: ABCD ∥gram,
BK ⊥ AD , AB ⊥ BD
AB=6, AK=3
Find: m∠A, BK
To proof
as given BK ⊥ AD
thus ΔABK is a right triangle.
thus by using the pythagoras theorem
we have
AB² = BK² + AK²
BK² = AB ²- AK²
= 6² - 3²
= 36 - 9
= 27
Hence
BK = <u> unit</u>
Now find the value m∠A
by using the trignometric function
FORMULA
cosA =
cosA =
=
m∠A = 60°
hence proved
Check out the graph below. We see that (3,0) is a root or x intercept. More specifically, the graph only touches the x axis at this point, rather than cross over (in contrast to a root like (-2,0) ). We say that the root here is of multiplicity 2. The multiplicity is due to the exponent of 2 over the (x-3) factor.
You can also say that the graph is decreasing on the left side of (3,0) and then it bounces off the root x = 3 to increase afterward. In this region, the graph never goes below the x axis.
