Question

The two shorter sides of a triangle measure $9$ and $18$. If the internal angle bisector drawn to the longest side measures $8$, find the measure of the longest side of the triangle.

• A. $18$
• B. $23$
• C. $22$
• D. $21$

Answer 1

The correct option is D $21$

let $\angle BAD$=$\theta$=$\angle DAC$ ( $\because$ Given)

$\cos \theta$=$\frac{(A^2+B^2-C^2)}{2AB}$

$\cos \angle BAD$=$\frac{(8^2+9^2-x^2)}{2\times 8\times 9}$ (in triangle ABD ) $\to$ 1

$\cos \angle CAD$=$\frac{\{{18}^2+8^2-{\left(2x\right)}^2\}}{2\times 18\times 8}$(in triangle ADC) $\to$ 2

from $\to$ 1 &$\to$ 2

$x=7$ (Substitute$\cos \angle BAD$ value in equation 2 )

$\therefore$ longest side of triangle will be =$x+2\times x=7+14=21$