Answer: ∠A= [tex]24\textdegree[/tex]
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,
[tex]\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[/tex]
Now, in ΔACD,
[tex]x+x=48\textdegree\\\\2x=48\textdegree\\\\\text{( Sum of interior anlges is equal to exterior angles )}\\\\x=\frac{48}{2}\\\\x=24\textdegree[/tex]
Hence , ∠A= [tex]24\textdegree[/tex]
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°
