Vertex of the angle is 11.31 cm far from the center of the circle.
Given in the question,
- Measure of angle formed by two tangents of the circle (∠PVQ) = 90°
- Measure of the radius (OP)= 8 cm
In ΔOPV and ΔOQV,
OP ≅ OQ (Radii of the circle)
PV ≅ QV (Tangents from a point outside the circle)
OV ≅ OV (Reflexive property)
ΔOPV ≅ ΔOQV (SSS property of congruence)
Therefore, m∠PVO = m∠QVO = 45°
By triangle sum theorem,
m∠POV + m∠PVO + m∠OPV = 180°
m∠POV + 45° + 90° = 180°
m∠POV = 45°
Hence, ΔOPV is an isosceles triangle.
OP = PV = 8 cm
By applying Pythagoras theorem in ΔOPV.
OV² = OP² + PV²
OV² = 8² + 8²
OV = √128
OV = 11.31 cm
Therefore, vertex V is 11.31 cm from the center of the circle.
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