In the figure below, cd bisects m∠acb, ab=bc, m∠bec=90°, and m∠dce=42°. find the measure of ∠a.

The measure of ∠A is 32°.

Step-by-step explanation:

Given : AB = BC

∠BEC=90°

∠DCE=42°

CD bisects ∠ACB :

∠ACD = ∠DCB = x

∠DCE= ∠BCE + ∠DCB = ∠BCE + x = 42°

To find : ∠A = ?

Solution:

InΔ ACE:

∠A +∠E +∠ACD + ∠DCB + ∠BCE = 180° (Angle sum property)

∠A +∠E +∠DCB + ∠DCB + ∠BCE = 180°

∠A +∠E +x + x + ∠BCE = 180°

∠A +∠E + x + 42° = 180°

∠A + x = 180° - 90° - 42° = 48°

∠A + x = 48°...[1]

In Δ ABC

AB =  BC (given)  

Hence, Isosceles triangle

∠A = ∠ACB = 2x ..[2]

(Angle opposite to equal sides are equal)

Using [2] in [1] we get:

2x + x = 48°

x = 16°

∠A  = 2x = 2 × 16° = 32°

the answer is (2x+8)²= 16

damm bro the struggle

step-by-step explanation: