Let triangle ABC be a triangle such that angle ACB is 135 degrees.
Prove that AB^2 = AC^2 + BC^2 - (Root 2) x AC x BC
To prove that AB^2 = AC^2 + BC^2 - (√2) x AC x BC, we can use the law of cosines.
The law of cosines states that for any triangle with sides a, b, and c, and the angle opposite side c denoted as θ, the following equation holds:
c^2 = a^2 + b^2 - 2ab cos(θ)
In the given triangle ABC, let side AB be c, side AC be a, and side BC be b. The angle opposite side AB is denoted as θ, which is equal to angle ACB.
Applying the law of cosines, we have:
AB^2 = AC^2 + BC^2 - 2AC * BC * cos(ACB)
Since we know that angle ACB is 135 degrees, we can substitute its value into the equation:
AB^2 = AC^2 + BC^2 - 2AC * BC * cos(135)
To simplify further, we need to determine the cosine of 135 degrees.
The cosine of 135 degrees can be determined using the unit circle or by using the concept of reference angles. The reference angle for 135 degrees is 45 degrees, which lies in the first quadrant.
In the first quadrant, the cosine function is positive. Therefore, cos(135) = cos(45) = 1/√2.
Thus, we can substitute cos(135) as 1/√2 into the equation:
AB^2 = AC^2 + BC^2 - 2AC * BC * (1/√2)
Simplifying:
AB^2 = AC^2 + BC^2 - (2/√2) * AC * BC
To simplify the term (2/√2), we can rationalize the denominator:
AB^2 = AC^2 + BC^2 - (√2) * AC * BC
Hence, we have proved that AB^2 = AC^2 + BC^2 - (√2) * AC * BC in triangle ABC with angle ACB as 135 degrees using the law of cosines.