Let l1-l2-l3 be the lengths of the internal bisectors of angles A- B- C respectively of a - ABC-Statement-1-cosA2l1-cosB2l2-cosC2l3-21a-1b-1cStatement-2-l21-bc-1-ab-c2-l22-ca-1-bc-a2- l32-ab-1-ca-b2-

L_1=2bcb+ccosA2...(1) ⇒cosA2l_1+cosB2l_2+cosC2l_3=(1a+1b+1c) Statement 1 is false. From equation (1), nbsp; l_1^2=4b^2c^2(b+c)^2cos^2A2 ⇒ l^2_1=bc[1 (ab+c) ...

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Byju's Answer

Standard IX

Mathematics

Any Point Equidistant from the End Points of a Segment Lies on the Perpendicular Bisector of the Segment

Let l1,l2,l3 ...

Question

Let $l_{1}, l_{2}, l_{3}$ be the lengths of the internal bisectors of angles $A, B, C$ respectively of a $Δ A B C .$

Statement-1; $\frac{cos \frac{A}{2}}{l_{1}} + \frac{cos \frac{B}{2}}{l_{2}} + \frac{cos \frac{C}{2}}{l_{3}} = 2 (\frac{1}{a} + \frac{1}{b} + \frac{1}{c})$

Statement-2: $l_{1}^{2} = b c [1 - {(\frac{a}{b + c})}^{2}]; l_{2}^{2} = c a [1 - {(\frac{b}{c + a})}^{2}]; l_{3}^{2} = a b [1 - {(\frac{c}{a + b})}^{2}]$

Statement-1 is True, Statement-2 is True; Statement-2 is a correct explanation for Statement -1

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Statement-1 is True, Statement-2 is True; Statement-2 NOT a correct explanation for Statement -1

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Statement-1 is True, Statement-2 is False

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Statement-1 is False, Statement-2 is True

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Solution

The correct option is D Statement-1 is False, Statement-2 is True
$l_{1} = \frac{2 b c}{b + c} cos \frac{A}{2} . . . (1)$

$\Rightarrow \frac{cos \frac{A}{2}}{l_{1}} + \frac{cos \frac{B}{2}}{l_{2}} + \frac{cos \frac{C}{2}}{l_{3}} = (\frac{1}{a} + \frac{1}{b} + \frac{1}{c})$
Statement-1 is false.

From equation $(1),$
$l_{1}^{2} = \frac{4 b^{2} c^{2}}{(b + c)^{2}} {cos}^{2} \frac{A}{2}$

$\Rightarrow l_{1}^{2} = b c [1 - {(\frac{a}{b + c})}^{2}]; l_{2}^{2} = c a [1 - {(\frac{b}{c + a})}^{2}];$
and $l_{3}^{2} = a b [1 - {(\frac{c}{a + b})}^{2}]$
Statement-2 is true.

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Similar questions

Match the equations A, B, C, D with the lines L₁, L₂, L₃, L₄, L₅ whose graphs are roughly drawn in the adjoining figure.

A: y = 2

B: y – 2 + 2 = 0

C: 3 + 2y = 6

D: y = -2

E : = 2

Q. AB, AC are tangents to a parabola

y^{2} = 4 a x

l_{1}, l_{2} . l_{3}

are the lengths of perpendiculars from A, B, C on any tangent to the parabola, then

Q. Let position vectors of points

A, B

and

C

of triangle

A B C

respectively be

^i +^j + 2^k,

^i + 2^j +^k

and

2^i +^j +^k .

Let

l_{1}, l_{2}, l_{3}

be the lengths of perpendiculars drawn from the orthocentre

O

on the sides

A B,

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and

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respectively. Then the value of

4 (l_{1} + l_{2} + l_{3})^{2}

a² + b² + c² - ab - bc - ca equals:

Show that the straight lines $L_{1} = (b + c) x + a y + 1 = 0, L_{2} = (c + a) x + b y + 1 = 0$ and $L_{3} = (a + b) x + c y + 1 = 0$ are concurrent.

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Any Point Equidistant from the End Points of a Segment Lies on the Perpendicular Bisector of the Segment

Standard IX Mathematics

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Let l1,l2,l3 be the lengths of the internal bisectors of angles A,B,C respectively of a ΔABC. Statement-1;cosA2l1+cosB2l2+cosC2l3=2(1a+1b+1c) Statement-2:l21=bc[1−(ab+c)2];l22=ca[1−(bc+a)2];l23=ab[1−(ca+b)2]

The correct option is D Statement-1 is False, Statement-2 is Truel1=2bcb+ccosA2...(1) ⇒cosA2l1+cosB2l2+cosC2l3=(1a+1b+1c) Statement-1 is false. From equation (1), l21=4b2c2(b+c)2cos2A2 ⇒l21=bc[1−(ab+c)2];l22=ca[1−(bc+a)2]; and l23=ab[1−(ca+b)2] Statement-2 is true.