The angles in triangle ABC satisfy 6sin∠A=3√3sin∠B=2√2sin∠C. If sin^2∠A=ab, where a and b are coprime positive integers, what is the value of a+b?
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If sin^2∠A=ab
its square there
can't you take that final step?
sinA = 8/25, so sin^2 A = ???
the answers wrong.........
To find the value of a + b, we need to determine the values of sin^2∠A, ∠A, ∠B, and ∠C in the given triangle.
Given: 6sin∠A = 3√3sin∠B = 2√2sin∠C
We can start by using the fact that the sum of the angles in a triangle is 180 degrees:
∠A + ∠B + ∠C = 180°
Now, let's focus on the given information:
6sin∠A = 3√3sin∠B = 2√2sin∠C
From this, we can see that the ratios of sin∠A, sin∠B, and sin∠C are in the ratio of 6:3√3:2√2.
To simplify this ratio, we divide every term by the smallest term (2√2):
(6/2√2)sin∠A = (3√3/2√2)sin∠B = sin∠C
Simplifying:
3√2sin∠A=3√3sin∠B=sin∠C
Now, we can assign a value to one of the angles. Let's choose ∠A:
Let x be the value of sin∠A. Then, we have:
3√2x = 3√3sin∠B = sin∠C
Since sin^2∠A = x^2, we need to find x^2.
To find x, we can use the fact that sin^2∠A + cos^2∠A = 1, which implies cos^2∠A = 1 - sin^2∠A:
cos^2∠A = 1 - x^2
Since cos^2∠A = cos^2(180° - ∠A) = cos^2∠C, we can rewrite the equation as:
cos^2∠C = 1 - x^2
We also know that sin^2∠C = (3√2x)^2:
sin^2∠C = 18x^2
Using the trigonometric identity sin^2∠C + cos^2∠C = 1, we can substitute the expressions for sin^2∠C and cos^2∠C:
18x^2 + 1 - x^2 = 1
Simplifying:
17x^2 = 0
The only possible value for x is 0 since it satisfies the equation. However, since we are looking for positive solutions, this is not a valid solution.
Hence, there is no positive solution for sin^2∠A = ab, and the value of a + b is undefined.