Question

In the figure below, CD bisects m∠ACB, AB=BC, m∠BEC=90°, and m∠DCE=42°. Find the measure of ∠A.

Answer 1

Answer:  ∠A= $24\textdegree$

Step-by-step explanation:

Since we have given that

CD bisects m∠ACB,

⇒  m∠ACD= m∠DCB =x (say)

And, AB=BC, m∠BEC=90°, and m∠DCE=42°

So,

m∠CAD=m∠ACD

In ΔDCE,

$\angle DCE+\angle CED+\angle CDE=180\textdegree\\\\(\text{Sum of anlges in triangle is }180\textdegree)\\\\42\textdegree+90\textdegree+\angle CDE=180\textdegree\\\\132\textdegree+\angle CDE=180\textdegree\\\\\angle CDE=180\textdegree-132\textdegree\\\\+\angle CDE=48\textdegree$

Now, in  ΔACD,

$x+x=48\textdegree\\\\2x=48\textdegree\\\\\text{( Sum of interior anlges is equal to exterior angles )}\\\\x=\frac{48}{2}\\\\x=24\textdegree$

Hence , ∠A= $24\textdegree$

Answer 2

Answer:

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°