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A circle is c...

Question

A circle is circumscribed about a triangle $A B C$ . A tangent is drawn to the circle through the point $B$ to intersect the extension of the side $C A$ beyond the point $A$ at the point $D$ . Let $P$ be the perimeter of triangle $A B C$ . Given $A B + A D = A C, C D = 3$ and $∠ B A C = 60^{\circ}$ then the value of $(3 - \sqrt{2}) P$ is

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Solution

The correct option is B 6
$B D^{2} = D A . D C = 3 x$
Using Cosine law on $∠ B A D$
In $Δ B A D,$ we get $x = 1$
Then applying Cosine law on $∠ B A C$ in $Δ B A C,$ we get $B C = \sqrt{3} \Rightarrow P = 3 + \sqrt{3}$

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A circle is circumscribed about a triangle ABC. A tangent is drawn to the circle through the point B to intersect the extension of the side CA beyond the point A at the point D. Let P be the perimeter of triangle ABC. Given AB+AD=AC, CD=3 and ∠BAC=60∘ then the value of (3−√2)P is

The correct option is B 6BD2=DA.DC=3x Using Cosine law on ∠BAD In ΔBAD, we get x=1 Then applying Cosine law on ∠BAC in ΔBAC, we get BC=√3⇒P=3+√3

The correct option is B 6
$B D^{2} = D A . D C = 3 x$
Using Cosine law on $∠ B A D$
In $Δ B A D,$ we get $x = 1$
Then applying Cosine law on $∠ B A C$ in $Δ B A C,$ we get $B C = \sqrt{3} \Rightarrow P = 3 + \sqrt{3}$