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 = [tex]3\sqrt{3}[/tex] unit
Now find the value m∠A
by using the trignometric function
FORMULA
cosA = [tex]\frac{BASE}{HYPOTENUSE}[/tex]
cosA = [tex]\frac{3}{6}[/tex]
= [tex]\frac{1}{2}[/tex]
m∠A = 60°
hence proved
