Question

Given: ABCD ∥gram, BK ⊥ AD , AB ⊥ BD AB=6, AK=3 Find: m∠A, BK, A

Answer 1

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 = $3\sqrt{3}$ unit

Now find the value  m∠A

by using the trignometric function

FORMULA

cosA = $\frac{BASE}{HYPOTENUSE}$

cosA = $\frac{3}{6}$

       = $\frac{1}{2}$

m∠A = 60°

hence proved