The centre O of a circle inside a triangle ABC is at a distance of 13 cm from each of the vertices of the triangle. The diameter of the circle is 10 cm and the circle touches only two sides of the triangle, AB and AC.
Consider the following statements: I. \(\angle ABC\) lies between 60° and 90°. II. If z is the distance in cm from the centre O of the circle to the midpoint of BC, then \(7\text{ cm} < z < 8\text{ cm}\). Which of the statements given above is/are correct?
I only
Since O is 13 cm from every vertex, O is the circumcentre with circumradius R = 13 cm. The circle of radius 5 cm touches AB and AC only, so the perpendicular distances from O to AB and AC both equal 5 cm, i.e. R cos C = R cos B = 5, giving cos B = cos C = 5/13 and hence B = C (triangle is isosceles with AB = AC). So sin B = 12/13, AB = AC = 2R sin B = 24 cm. Also A = 180 deg - 2B, so sin A = 2 sin B cos B = 120/169 and cos A = 1 - 2cos^2B = 119/169. Angle \(B = \cos^{-1}(5/13) \approx 67.4^\circ\), which lies between 60° and 90°, so statement I is true. The distance from O to the midpoint of BC is \(R\cos A = 13 \times \tfrac{119}{169} = \tfrac{119}{13} \approx 9.15\text{ cm}\), which is NOT between 7 cm and 8 cm, so statement II is false. Hence only statement I is correct.
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