Question

With usual notation, if in a triangle $ABC,$ $a+3=b+2=11$ and $\cos C=\frac{2}{3},$ then

• A. Length of the tangent from the vertex $C$ to the circle escribed to side $AB$ is $12.$
• B. Length of the tangent from the vertex $C$ to the incircle of the triangle is $4.$
• C. Sum of lengths of tangents from the vertices $A,B,C$ to the circle escribed to the side $AB$ is $38.$
• D. Sum of lengths of tangents from the vertices $A,B,C$ to the incircle of the triangle is $24.$

Answer 1

Answer: (A), (C), (D)

Sum of lengths of tangents from the vertices $A,B,C$ to the incircle of the triangle is $24.$

Given, $a+3=11$ and $b+2=11$
$\therefore a=8;b=9$
$\cos C=\frac{2}{3}$
$\Rightarrow \frac{a^2+b^2-c^2}{2ab}=\frac{2}{3}$
$\Rightarrow \frac{64+81-c^2}{2\cdot 8\cdot 9}=\frac{2}{3}$
$\Rightarrow 2\times 144=3(145-c^2)$
$\Rightarrow 2\times 48=145-c^2$
$\Rightarrow 96=145-c^2$
$\Rightarrow c^2=145-96$
$\Rightarrow c^2=49$
$\therefore c=7$

$CE=CF$
$\Rightarrow 9+x=8+7-x$
$\Rightarrow 2x=6\Rightarrow x=3$
Hence, length of tangent from $C$ to $S_1$ is $9+x=8+7-x=12$

Sum of the lengths of tangents from $A,B,C$ to $S_1$ is
$2x+2(7-x)+(9+x)+(8+7-x)=38$

Length of tangent from $C$ to the inscribed circle is $s-c=12-7=5$

Sum of the lengths of tangents from $A,B,C$ to the inscribed circle is $a+b+c=24$