[tex] \underline{ \underline{ \text{Question}}} : [/tex]
In the adjoining figure , ABC is an isosceles triangle. BO and CO are the bisectors of [tex] \angle[/tex] ABC and [tex] \angle[/tex] ACB respectively. Prove that BOC is an isosceles triangle.
~Thanks in advance !
![[tex] \underline{ \underline{ \text{Question}}} : [/tex]In The Adjoining Figure , ABC Is An Isosceles](https://brainimage.s3.us-east-005.backblazeb2.com/contents/attachments/9IR3XgX79oxTHAmOAL8xQtxS8ZMKdSbS.jpeg)
Answers
Here it's given that ∆ABC is a isosceles triangle. We know that in a isosceles triangle opposite angles are equal. And the angles opposite to equal sides are also equal.
Hence here ,
AB = AC ∠ ABC = ∠ACB .Figure :-
[tex] \setlength{\unitlength}{1 cm}\begin{picture}(12,12)\put(0,0){\line(1,0){4}}\put(4,0){\line(-1,2){2}}\put(0,0.001){\line(1,2){2}}\put(0,0.01){\line(1,1){2}}\put(3.99,0){\line(-1,1){2}}\put(0,-0.3){$\bf B $}\put(4,-0.3){$\bf C$}\put(2,4.2){$\bf A$}
\put(1.8,2.2){$\bf o$}\end{picture}[/tex]
If LaTeX doesn't work on app kindly see the attachment .
Here we will use a theorem which is ,
[tex]\large\underline{\underline{\textsf{\textbf{\red{\blue{$\leadsto$} Theorem :- }}}}}[/tex]
Sides opposite to equal angles are equal.Hence here,
[tex]\tt:\implies \angle ABC =\angle ACB \\\\\tt:\implies \dfrac{1}{2}\times \angle ABC = \dfrac{1}{2}\times \angle ACB \\\\\tt:\implies \boxed{\bf \blue{\angle OBC = \angle OCB} }[/tex]
Now in ∆OBC , we can see that ∠OBC = ∠OCB. (just proved ) . Hence we can say that sides OB and OC are equal. Since the sides opposite to equal angles are equal. Therefore in ∆OBC , two sides are equal.
Therefore ∆ OBC is an isosceles triangle.
Hence Proved !![[tex] \underline{ \underline{ \text{Question}}} : [/tex]In The Adjoining Figure , ABC Is An Isosceles](https://brainimage.s3.us-east-005.backblazeb2.com/answers/attachments/NT2TEI3wINzzSNqlJFTL1hvxf64h9TmI.jpeg)
Answer:
See Below.
Step-by-step explanation:
We are given the isosceles triangle ΔABC. By the definition of isosceles triangles, this means that ∠ABC = ∠ACB.
Segments BO and CO bisects ∠ABC and ∠ACB.
And we want to prove that ΔBOC is an isosceles triangle.
Since BO and CO are the angle bisectors of ∠ABC and ∠ACB, respectively, it means that ∠ABO = ∠CBO and ∠ACO = ∠BCO.
And since ∠ABC = ∠ACB, this implies that:
∠ABO = ∠CBO =∠ACO = ∠BCO.
This is shown in the figure as each angle having only one tick mark, meaning that they are congruent.
So, we know that:
[tex]\angle ABC=\angle ACB[/tex]
∠ABC is the sum of the angles ∠ABO and ∠CBO. Likewise, ∠ACB is the sum of the angles ∠ACO and ∠BCO. Hence:
[tex]\angle ABO+\angle CBO =\angle ACO+\angle BCO[/tex]
Since ∠ABO =∠ACO, by substitution:
[tex]\angle ABO+\angle CBO =\angle ABO+\angle BCO[/tex]
Subtracting ∠ABO from both sides produces:
[tex]\angle CBO=\angle BCO[/tex]
So, we've proven that the two angles are congruent, thereby proving that ΔBOC is indeed an isosceles triangle.
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Step-by-step explanation:
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Answers
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Here given that,
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(1)
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Answers
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How does transformation of a function happens?The transformation of a function may involve any change.
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Let's put them vertically because that helps me explain this
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Step-by-step explanation:
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Answers
Answer:
The proof is shown below.
Step-by-step explanation:
Given: B is mid point of AC and C is mid-point of BD.
To prove: AD=3AB
It is given that B is mid-point of AC.
⇒ AB = BC ...(1)
Also, C is mid-point of BD.
⇒ BC = CD ...(2)
From (1) and (2),
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Answers
Step-by-step explanation:
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Answers
The value of p and q in the right triangle are as follows:
p ≈ 30.6 units
q ≈ 14.9 units
How to find the side of a right triangle?A right triangle is a triangle that has one of its angles as 90 degrees. The sum of angles in a triangle is 180 degrees.
Therefore, the side of a right triangle can be found using trigonometric ratios as follows:
Let's find p and q using the trigonometric ratios.
Hence,
sin 64 = opposite / hypotenuse
opposite side = p
hypotenuse = 34
Therefore,
sin 64 = p / 34
cross multiply
p = 34 sin 64
p = 30.5589975742
p ≈ 30.6 units
Hence,
cos 64 = adjacent / hypotenuse
cos 64 = q / 34
cross multiply
q = 34 cos 64
q = 14.9046189908
q = 14.9 units
Therefore,
p ≈ 30.6 units
q ≈ 14.9 units
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Q.2*
Simplify
(2a2 – 4a – 8)(a + 1)-1
A. 2a - 6
2
a+1
B. 2a2 - 6a - 2
2
C. 2a + 6+
a+1
Answers
i think the B is the answer