To find the value of [tex]\( \angle BAC \)[/tex] in the triangle [tex]\( \triangle ABC \)[/tex] given that [tex]\( AB = 18 \)[/tex] cm, [tex]\( AC = 20 \)[/tex] cm, and the area of the triangle is [tex]\( 90 \sqrt{3} \)[/tex] cm², follow these steps:
1. Formula for area of a triangle:
The area of a triangle can be calculated using two sides and the included angle as follows:
[tex]\[
\text{Area} = \frac{1}{2} \times AB \times AC \times \sin(\angle BAC)
\][/tex]
2. Substitute the given values:
[tex]\[
90 \sqrt{3} = \frac{1}{2} \times 18 \times 20 \times \sin(\angle BAC)
\][/tex]
3. Simplify the equation:
[tex]\[
90 \sqrt{3} = 180 \times \sin(\angle BAC)
\][/tex]
4. Isolate [tex]\( \sin(\angle BAC) \)[/tex]:
[tex]\[
\sin(\angle BAC) = \frac{90 \sqrt{3}}{180}
\][/tex]
5. Simplify the fraction:
[tex]\[
\sin(\angle BAC) = \frac{\sqrt{3}}{2}
\][/tex]
6. Find the angle [tex]\( \angle BAC \)[/tex] using the inverse sine function:
The value of [tex]\( \sin(\theta) = \frac{\sqrt{3}}{2} \)[/tex] corresponds to the angle [tex]\( \theta = 60^\circ \)[/tex].
Therefore, the value of [tex]\( \angle BAC \)[/tex] is:
[tex]\[
\angle BAC = 60^\circ
\][/tex]
This should be the accurate measure for [tex]\( \angle BAC \)[/tex] in the given triangle.
