In a △ABC,ㄥABC = ㄥACB and the bisectors of ㄥABC and ㄥACB intersect at O such that ㄥBOC = 120⋄. Show that ㄥA = ㄥB = ㄥC = 60⋄. plsssss give the answer with proper and explained steps. And whosoever will give proper reasons I will mark that person brainliest, like his answer and will raye his answer with full 5 stars, with 30 points. and no, spams, or wrong answers, pls don't do that or I will report !!!...guys plssssssssss help...!!! !!!...plssss help fast...!!! :(
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7th
Maths
The Triangle and Its Properties
Angle Sum Property
In the given figure, BO, CO...
MATHS
In the given figure, BO, CO are the angle bisectors of external angles of ΔABC. Then ∠BOC is __________.
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ANSWER
As BO and CO are the angle bisectors of external angles of△ABC, Then
∠1=∠2
∠4=∠3
We know, ∠A+∠ABC+∠ACB=180
∘
…eqn(1)
And ∠ABC=180−2∠1
∠ACB=180−2∠4
Putting it in the eqn (1), we get
∠A+180−2∠1+180−2∠4=180
⇒∠1+∠4=90+
2
1
∠A…eqn(2)
Also we know from the figure, ∠BOC+∠1+∠4=180
∘
∠BOC=180−∠1−∠4
From eqn (2)
∠BOC=180−90−
2
1
∠A
⇒∠BOC=90
∘
−
2
1
∠A
Answer:
Given :
A △ABC such that the bisectors of ∠ABC and ∠ACB meet at a point O.
To prove :
∠BOC=90
o
+
2
1
∠A
Proof :
In △BOC,
∠1+∠2+∠BOC=180
o
In △ABC,
∠A+∠B+∠C=180
o
∠A+2(∠1)+2(∠2)=180
o
2
∠A
+∠1+∠2=90
o
∠1+∠2=90
o
−
2
∠A
Therefore,
90
o
−
2
∠A
+∠BOC=180
o
∠BOC=90
o
+
2
∠A