Answer:
<u>y = w and ΔABC ~ ΔCDE</u>
Step-by-step explanation:
Given sin(y°) = cos(x°)
So, ∠y + ∠x = 90° ⇒(1)
And as shown at the graph:
ΔABC is aright triangle at B
So, ∠y + ∠z = 90° ⇒(2)
From (1) and (2)
<u>∴ ∠x = ∠z </u>
ΔCDE is aright triangle at D
So, ∠x + ∠w = 90° ⇒(3)
From (1) and (3)
<u>∴ ∠y = ∠w</u>
So, for the triangles ΔABC and ΔCDE
- ∠A = ∠C ⇒ proved by ∠y = ∠w
- ∠B = ∠D ⇒ Given ∠B and ∠D are right angles.
- ∠C = ∠E ⇒ proved by ∠x = ∠z
So, from the previous ΔABC ~ ΔCDE by AAA postulate.
So, the answer is <u>y = w and ΔABC ~ ΔCDE</u>
They are all vertical angles.
Answer:
though this looks difficult. It's easier than you think. Just not visible until its visible. If I knew the answer, i would have told you... sorry.
