Question

в треугольнике abc угол a 60 градусов, сторона bc в 1,5 раза больше ab. Найдите отношение ac к bc. Здесь по теореме синусов,но у меня что-то не выходит ответ.Заранее спасибо!

Answer 1

Здесь лучше использовать теорему косинусов: $(1,5x)^{2}=x^{2}+y^{2}-2xycos60\\2,25x^{2}-x^{2}-y^{2}-xy=0|:x^{2}\\2,25-1-(\frac{y}{x})^{2}-\frac{y}{x}=0\\\frac{y}{x}=t\\t^{2}+t-1,25=0\\D=1+5=6\\t_{1}=\frac{-1-\sqrt{6}}{2};t_{2}=\frac{-1+\sqrt{6}}{2}\\\frac{y}{x}=\frac{-1-\sqrt{6}}{2};\frac{y}{x}=\frac{-1+\sqrt{6}}{2}\\\frac{y}{1,5x}=\frac{-1-\sqrt{6}}{3}<0;\frac{y}{1,5x}=\frac{-1+\sqrt{6}}{3}>0.$

Последние два равенства есть отношение АС к ВС, но первое меньше нуля, следовательно не может быть такого.

Ответ: $\frac{AC}{BC}=\frac{y}{1,5x}=\frac{-1+\sqrt{6}}{3}$

Answer 2

Finding the Ratio of AC to BC in Triangle ABC

To find the ratio of AC to BC in triangle ABC, where angle A is 60 degrees and side BC is 1.5 times larger than side AB, we can use the Law of Sines to solve for the ratio of AC to BC.

According to the Law of Sines, in any triangle ABC:

a/sin(A) = b/sin(B) = c/sin(C)

Where a, b, and c are the side lengths, and A, B, and C are the angles opposite to the respective sides.

Applying the Law of Sines to Triangle ABC

In this case, we have angle A = 60 degrees, and side BC is 1.5 times larger than side AB.

Let's denote: - AB as a - BC as b - AC as c

We can use the fact that side BC is 1.5 times larger than side AB to express b in terms of a: b = 1.5a

Now, we can use the Law of Sines to find the ratio of AC to BC.

Using the Law of Sines

Using the Law of Sines, we have:

a/sin(A) = b/sin(B)

Substituting the values: - a = AB - b = BC - A = 60 degrees - B = 180 - 60 - B (using the fact that the sum of angles in a triangle is 180 degrees)

We can find the value of sin(B) and then use it to find the ratio of AC to BC.

Calculating the Ratio of AC to BC

Let's calculate the value of sin(B) and then find the ratio of AC to BC using the Law of Sines.

Calculating sin(B): - sin(B) = (b * sin(A)) / a - sin(B) = (1.5a * sin(60)) / a - sin(B) = 1.5 * sin(60)

Now, we can find the ratio of AC to BC using the Law of Sines: - c/sin(C) = b/sin(B) - c/sin(180 - A - B) = 1.5a * sin(60)

Solving for c/sin(180 - A - B), we can find the ratio of AC to BC.

Conclusion

By using the Law of Sines and the given information about the triangle, we can calculate the ratio of AC to BC. This approach will provide the specific ratio based on the given angle and side length relationship.