Γ is the circumcircle of triangle ABC with radius 12√2. If 9cos(∠B+∠C)cos∠A=−1, what is the value of BC?
To find the value of BC, we need to use the given information about the circumcircle and the equation involving angles A, B, and C.
First, let's understand what the equation 9cos(∠B+∠C)cos∠A = -1 means. The sum of angles B and C is equal to ∠B + ∠C = 180° - ∠A. Also, we know that cos(180° - ∠A) = -cos∠A. Therefore, the equation can be rewritten as:
9cos(∠B+∠C)cos∠A = -1
9*(-cos∠A)*cos∠A = -1
-9cos^2∠A = -1
Now, let's solve this equation for the value of cos^2∠A:
-9cos^2∠A = -1
cos^2∠A = 1/9
Taking the square root of both sides, we get:
cos∠A = ±1/3
Since cos∠A is positive, we take the positive root:
cos∠A = 1/3
Now that we know the cosine of angle A, we can use it to find the length of side BC. In a triangle inscribed in a circle, the Law of Cosines can be used to relate the side lengths and angles:
c^2 = a^2 + b^2 - 2ab*cosC,
where c is the side opposite angle C, and a and b are the sides opposite angles A and B, respectively. In our case, side BC is opposite angle A. Therefore, we have:
BC^2 = AB^2 + AC^2 - 2AB*AC*cos∠A.
Since the triangle is inscribed in the circumcircle, we know that AB = AC = radius = 12√2. Plugging in all the known values, we get:
BC^2 = (12√2)^2 + (12√2)^2 - 2(12√2)(12√2)(1/3)
BC^2 = 576 + 576 - 2*144
BC^2 = 1152 - 288
BC^2 = 864.
Taking the square root of both sides, we find:
BC = √864 = 12√6.
So, the value of BC is 12√6.