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  • 3 answers
O .1

Kushi P Shetty 2 years ago

Which and all expected or sure question for board exam for 2023

Ashutosh Tiwari 2 years ago

Ap
  • 1 answers

Preeti Dabral 2 years ago

To show that the perimeter of AABL is constant, we need to prove that AB + BA + AL + BL is constant.

First, let us label the angles in the figure as follows:

∠NLM = ∠LNM = α (since LM and LN are tangents from an external point, they are equal in measure)
∠MAB = β (angle between the tangent AB and the chord MC)
∠NBA = β (angle between the tangent AB and the chord NC)
∠ALN = γ (angle between the tangent LN and the chord MN)
∠BLM = γ (angle between the tangent LM and the chord MN)

Using the fact that the sum of angles in a triangle is 180 degrees, we can write:

∠LAM = 180 - α - β
∠LBN = 180 - α - β

Now, let us consider the perimeter of AABL:

AB + BA + AL + BL

Using the fact that angles in the same segment of a circle are equal, we can write:

∠MCN = ∠MAB + ∠NBA = 2β

Using the fact that the opposite angles in a cyclic quadrilateral add up to 180 degrees, we can write:

∠ALB = 180 - ∠MCN = 180 - 2β

Now, let us consider the triangles LAM and LBN. Using the fact that angles in a triangle add up to 180 degrees, we can write:

∠LAM + ∠ALN + ∠ALM = 180
∠LBN + ∠BLM + ∠BLN = 180

Substituting the values we have calculated for ∠LAM and ∠LBN, and using the fact that ∠ALN = ∠BLM = γ, we get:

(180 - α - β) + γ + ∠ALM = 180
(180 - α - β) + γ + ∠BLN = 180

Simplifying, we get:

∠ALM = α + β - γ
∠BLN = α + β - γ

Finally, let us consider the perimeter of AABL again:

AB + BA + AL + BL = AB + BA + 2AL sin γ + 2BL sin γ

Using the fact that AL = AM and BL = BN, and using the sine rule in triangles LAM and LBN, we get:

AB + BA + 2AM sin γ + 2BN sin γ = AB + BA + 2LM sin (α + β - γ) + 2LN sin (α + β - γ)

Using the fact that LM = LN (since they are tangents from an external point), we get:

AB + BA + 2LM sin (α + β - γ) + 2LM sin (α + β - γ)

Simplifying, we get:

AB + BA + 4LM sin (α + β - γ)

Using the fact that sin (α + β - γ) = sin (180 - γ) = sin γ, we get:

AB + BA + 4LM sin γ

Substituting the value of LM (which is equal to LN), we get:

AB + BA + 4LN sin γ

But we know that sin γ is a constant, since it is determined by the position of point L and the circle MN. Therefore, the perimeter of AABL is constant, since it does not depend on the position of point C on the arc MN.

Hence, we have proved that the perimeter of AABL is constant.

  • 3 answers

Arshad Alam 2 years ago

Every composite number can be expanded in the product of prime no. Such tha p and q

Aman Baghel 2 years ago

Every composite number can be written in the product of prime no.

Ashish Ojha 2 years ago

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  • 2 answers

Riya Lama 2 years ago

Send karte hu

Riya Lama 2 years ago

This content has been hidden. One or more users have flagged this content as inappropriate. Once content is flagged, it is hidden from users and is reviewed by myCBSEguide team against our Community Guidelines. If content is found in violation, the user posting this content will be banned for 30 days from using Homework help section. Suspended users will receive error while adding question or answer. Question comments have also been disabled. Read community guidelines at https://mycbseguide.com/community-guidelines.html

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  • 3 answers

Soniya Pandey 2 years ago

Imp h bhut

Wellz Sari 2 years ago

ha aur bahat imp hai
Yes
  • 5 answers

Wellz Sari 2 years ago

No

Shanu Yadav 2 years ago

Noo

Srishti Thakur 2 years ago

No you have to only write the correct options of mcqs
That means no..
Step by step solution is not required in MCQ type questions. Only tick/write down the correct option. Writing a complete solution is nothing but wastage of time in board.
  • 2 answers

Omya Aryan 2 years ago

Nth term = a+(n-1)d… a+(n-1)2a… a+2an-2a… 2an-a… (2n-1)a… option (b)
Option no. (b) (2n-1)a
  • 2 answers

Preeti Dabral 2 years ago

(6q+1)(6q+2)(6q+3)

in all 6q is common

then 6q is out

=6q(1+2+3)

=6q(6)

=36q/6

=6q

Yes, this is divisible by 6.

_Jass_ Mahey_ 2 years ago

6q
  • 3 answers

Omya Aryan 2 years ago

Not in syllabus
By Euclid's division algorithm, a=bq+r Take b=4 Since 0 < r < 4, r = 0,1,2,3 So, a=4q,4q+1,4q+2,4q+3 Clearly, a=4q,4q+2 are even, as they are divisible by 2. Therefore 'a' cannot be 4q,4q+2 as a is odd. But 4q+1,4q+3 are odd, as they are not divisible by 2. Any positive odd integer is of the form (4q+1) or (4q+3).

Abhi® Rana 2 years ago

Hlo....HINDI...M PUCHO TO BTA SKTE H.
  • 1 answers

Shristi Kumari 2 years ago

7×-15y=2 eq 1 ×+2y=3 eq 2 From eq 1 7×=2+15y ×=2+15y apon 7 Answer
  • 2 answers

_Jass_ Mahey_ 2 years ago

Pair linear equations in two variables

Karan Raj 2 years ago

Pair of linear equation
  • 1 answers

Ankit Class 2 years ago

Let theta be angle A Cosec A =√5 Cot A- cos A CosA/sinA -cosA CosA(cosecA-1) SinA = 1/√5 CosA =√1-sin^2 A CosA= √ 1-1/5 CosA=√4/5 CosA=2/√5 Now CotA-cosA=cosA(cosecA-1) CotA-cosA=2/√5(√5-1)
  • 3 answers

Hardik Kamble 2 years ago

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Karan Kundar 2 years ago

This content has been hidden. One or more users have flagged this content as inappropriate. Once content is flagged, it is hidden from users and is reviewed by myCBSEguide team against our Community Guidelines. If content is found in violation, the user posting this content will be banned for 30 days from using Homework help section. Suspended users will receive error while adding question or answer. Question comments have also been disabled. Read community guidelines at https://mycbseguide.com/community-guidelines.html

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  • Don't post personal information, mobile numbers and other details.
  • Don't use this platform for chatting, social networking and making friends. This platform is meant only for asking subject specific and study related questions.
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Remember the goal of this website is to share knowledge and learn from each other. Ask questions and help others by answering questions.

Pawan Gambhir 2 years ago

Ncert question
  • 3 answers
4/x + 9/x + 10 = 1/2 20x + 26 = x 20x + 26 - 26 = x - 26 20x = x-26 20x-x=x-26-x 19x=x-26-x 19x=-26 19x/19=26/19 x=-26/19 Answer...

Pooja Kumari 2 years ago

-26/19

Ankush Pahwa 2 years ago

LHS TAKING LCM x WE GET,, 4+9+10x/x =1/2 13+10x/x =1/2 CROSS MULTIPLICATION LHS & RHS WE GET,,, 26+20x = x 26= x-20x 26= -19x x = 26/-19 ANSWER
  • 3 answers
+ - is equal to -

Harshita Saini 2 years ago

-

Abhi® Rana 2 years ago

Hlo .....guys ...best of luck 21 march matmatic exam and plz bless me....HAR HAR MHADEV ............AGAR KISI KO BHEE BAAT KRNE H TO MUJE CALL KR SKTA H OR HM SATH M MATH K EXAMPLE .SOLUTION ETC KRENGE ....NUMBER TBHI MILEGA JN REPLY M KOI MESSAGE AAYEGA ......THANKU...💜
  • 1 answers

Aditi Chaudhary 2 years ago

x2+(y-3V2x)2=1 is a equation of an ellipse in the coordinates plane. The equation of an ellipse is typically given in a form of (x^2)/a^2+(y^2)/b^2 = 1, where a and b are the lengths of a semi major and semi minar axes, respectively. In the given equation , the semi major axis is 1 and semi minar axis is sqrt(3)
  • 1 answers
LHS = (1-Sin²A/Sin A)(1-Cos²A/Cos A)=SinA CosA=1/2Sin2A RHS=1/TanA+CotA=TanA/1+Tan²A=1/2Sin²A LHS=RHS [Hence proved]

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