Vectors, projections, and a suspect angle
In the plane, points A(1, 2), B(5, 4), and C(3, 7) are given. A student computes the angle at B using vectors BA and BC, obtains about 53 degrees, and then claims the triangle is acute. A second student computes the area via the cross product and gets 10. They then compute the circumradius using R = abc/(4K) and get a value that seems too large relative to the longest side. The pair must audit each computation, decide whether the triangle is acute, and reconcile the circumradius with the geometry.
1. Compute vectors BA and BC, their lengths, the dot product, and the angle at B. Judge the 53-degree claim.
Read the worked answer
BA = A - B = (-4, -2), |BA| = sqrt(20) = 2sqrt5 ≈ 4.472. BC = C - B = (-2, 3), |BC| = sqrt(13) ≈ 3.606.
Dot product: (-4)(-2) + (-2)(3) = 8 - 6 = 2.
cos(angle) = 2/(sqrt20 * sqrt13) = 2/sqrt260 ≈ 2/16.125 ≈ 0.1240, so the angle ≈ 82.9 degrees, not 53.
The student's figure is wrong. Since the dot product is positive, the angle at B is acute, but it is close to 83 degrees, nearly a right angle.
2. Determine whether the triangle is acute, right, or obtuse by checking all three angles, and verify the area of 10.
Read the worked answer
Sides: AB = sqrt20, BC = sqrt13, CA = A - C = (-2, -5), |CA| = sqrt29.
Largest side squared: 29. Sum of other squares: 20 + 13 = 33 > 29, so the angle opposite CA (angle at B) is acute. Check angle at A: sides AB^2 = 20 versus BC^2 + CA^2 = 13 + 29 = 42 > 20, acute. Angle at C: 13 versus 20 + 29 = 49, acute. All acute.
Area: 0.5|BA x BC| = 0.5|(-4)(3) - (-2)(-2)| = 0.5|-12 - 4| = 0.5(16) = 8, not 10.
So the area claim is also wrong; the correct area is 8.
3. Using the corrected side lengths and area, compute the circumradius and explain the student's impression that it was 'too large'.
Read the worked answer
R = abc/(4K) with a = sqrt13, b = sqrt29, c = sqrt20, K = 8.
abc = sqrt(132920) = sqrt(7540) ≈ 86.83. 4K = 32. R ≈ 2.713.
Longest side is sqrt29 ≈ 5.385, and R = 2.71 is about half of it, which is normal: for any triangle the circumradius is at least half the longest side (equality only for a right triangle with that side as hypotenuse). Here 2.71 > 2.69, barely, consistent with an angle just under 90 degrees at B.
The student's 'too large' impression came from using K = 10, which shrinks R to about 2.17, below the half-longest-side bound - an impossible value that should have flagged the area error.