Question

In \triangle ABC, we have \angle BAC = 60^\circ and \angle ABC = 45^\circ. The bisector of \angle A intersects \overline{BC} at point T, and AT = 24. What is the area of \triangle ABC?

Answer 1

Answer:

340.63

Step-by-step explanation:

To find the area we need to calculate the height of the triangle and one of its sides. the bisector cuts the 60° angle in two of 30°. In the picture I drew the triangle and the sides we need to calculate are y and h.

$\angle C = 180-45-60= 75.$

$\angle ACT = 180-30-75 = 75.$

We are going to calculate x, y and h with sin law:

$\frac{x}{sin(75)}= \frac{24}{sin(75)}$

$x= \frac{24*sin(75)}{sin(75)}=24.$

$\frac{y}{sin(60)}= \frac{24}{sin(45)}$

$y= \frac{24*sin(60)}{sin(75)}=29.39.$

$\frac{h}{sin(75)}= \frac{24}{sin(90)}$

$h= \frac{24*sin(75)}{sin(90)}=23.18.$

Then, the area of the triangle is

A = $\frac{y*h}{2} = \frac{29.39*23.18}{2}= 340.63.$