In a quadrilateral ABCD - AC is the bisector the angle between AB and AD which is 2 -3 - If 15- AC -3- AB -5- AD - then cosine of the angle between BA and DC is correct answer -1- wrong answer -0-25 A- -1-7B- 2-7-1-7-72D- -2-7

The correct option is B 2√(7)Given : nbsp;15|AC| = 3|AB| = 5|AD| Let |AC| =x, we get |AB| = 5x, |AD| =3x Drawing CE is parallel to AB, nbsp; Now, nbsp; ...

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Byju's Answer

Standard XII

Mathematics

Condition of Concurrency of 3 Straight Lines

In a quadrila...

Question

In a quadrilateral $A B C D$ , $- - \to A C$ is the bisector the angle between $- - \to A B$ and $- - \to A D$ which is $\frac{2 π}{3}$ . If $15 | - - \to A C | = 3 | - - \to A B | = 5 | - - \to A D |$ , then cosine of the angle between $- - \to B A$ and $- - \to D C$ is
(correct answer + 1, wrong answer - 0.25)

- \frac{1}{\sqrt{7}}

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\frac{1}{\sqrt{7}}

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\frac{2}{\sqrt{7}}

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- \frac{2}{\sqrt{7}}

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Solution

The correct option is C $\frac{2}{\sqrt{7}}$
Given : $15 | - - \to A C | = 3 | - - \to A B | = 5 | - - \to A D |$
Let $| - - \to A C | = x$ , we get
$| - - \to A B | = 5 x, | - - \to A D | = 3 x$

Drawing $C E$ is parallel to $A B$ ,
Now,
in $Δ D E C,$
$\frac{sin (60 - θ)}{x} = \frac{sin θ}{2 x} \Rightarrow tan θ = \frac{\sqrt{3}}{2} \Rightarrow cos θ = \frac{2}{\sqrt{7}}$

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In a quadrilateral ABCD, −−→AC is the bisector the angle between −−→AB and −−→AD which is 2π3. If 15|−−→AC|=3|−−→AB|=5|−−→AD|, then cosine of the angle between −−→BA and −−→DC is ​​​​​​​(correct answer + 1, wrong answer - 0.25)

The correct option is C 2√7Given : 15|−−→AC|=3|−−→AB|=5|−−→AD| Let |−−→AC|=x, we get |−−→AB|=5x,|−−→AD|=3x Drawing CE is parallel to AB, Now, in ΔDEC, sin(60−θ)x=sinθ2x⇒tanθ=√32⇒cosθ=2√7