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olga2289 [7]
3 years ago
13

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?

Mathematics
1 answer:
Aleks [24]3 years ago
4 0

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.

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