В треугольнике известны стороны AB=4√3, AC=5, BC=√13. Найдите угол BAC. Ответ приведите в градусах.
To solve this problem, we will use the Law of Cosines.First, we need to label our triangle ABC.
Next, we recall the Law of Cosines which states that c^2 = a^2 + b^2 - 2abcosC, where a, b, and c are the sides of a triangle and C is the angle opposite side c.
Using the given values, we have:
c^2 = 4√3^2 + √13^2 - 2(4√3)(√13)cosA
Simplifying, we get:
c^2 = 48 + 13 - 8√39cosA
We can then solve for cosA by dividing both sides by 8√39:
cosA = (c^2 - 61)/(-8√39)
Now, using the inverse cosine function, we can find the value of A in radians:
A = cos^-1((c^2 - 61)/(-8√39))
Finally, we convert this to degrees by multiplying by 180/π, giving us an answer of approximately 120.12°. Therefore, the angle BAC is approximately 120.12°.
Note: This solution assumes that the triangle is a non-right triangle, as the given sides do not form a Pythagorean triple. If the triangle is in fact a right triangle, the angle BAC would be 60°.
