Digital SAT Math Practice Question #1199

⚡ **Hack the Test: Leverage Geometric Properties & Common Triangles** 1. **Identify Key Geometric Features:** The points \(O(0,0)\), \(A(8,0)\), and \(B(0,6)\) immediately tell you that angle \(AOB\) is a right angle (\(90^\circ\)). Inscribed angles that are \(90^\circ\) subtend a diameter. Therefore, \(AB\) is the diameter of the circle. 2. **Find Point P Quickly:** The center of the circle is the midpoint of \(AB\), which is \(( (8+0)/2, (0+6)/2 ) = (4,3)\). The radius is half the length of \(AB\), so \(r = \frac{\sqrt{8^2 + 6^2}}{2} = \frac{\sqrt{64+36}}{2} = \frac{10}{2} = 5\). The equation of the circle is \((x-4)^2 + (y-3)^2 = 25\). Point \(P\) has an \(x\)-coordinate of 8. Substitute \(x=8\): \((8-4)^2 + (y-3)^2 = 25 \Rightarrow 4^2 + (y-3)^2 = 25 \Rightarrow 16 + (y-3)^2 = 25 \Rightarrow (y-3)^2 = 9 \Rightarrow y-3 = \pm 3\). Since \(y eq 0\), we take \(y-3=3\), so \(y=6\). Thus, \(P=(8,6)\). 3. **Recognize the Right Triangle:** Now you have \(O(0,0)\), \(P(8,6)\), and \(A(8,0)\). Notice that points \(P\) and \(A\) share the same \(x\)-coordinate (8). This means the line segment \(PA\) is vertical. Since \(OA\) lies on the x-axis (horizontal), angle \(PAO\) is a right angle (\(90^\circ\)). Triangle \(OPA\) is a right-angled triangle. 4. **Use Trigonometry for Common Angles:** In right triangle \(OPA\): * Length of \(PA = |6-0| = 6\). * Length of \(OA = |8-0| = 8\). * Length of \(OP = \sqrt{8^2+6^2} = 10\) (a 6-8-10 triangle, which is a 3-4-5 triangle scaled by 2). We need angle \(OPA\). The side opposite to \(\angle OPA\) is \(OA=8\), and the side adjacent is \(PA=6\). \(\tan(\angle OPA) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{OA}{PA} = \frac{8}{6} = \frac{4}{3}\). Recognize that \(\arctan(4/3)\) is a common angle in 3-4-5 right triangles, approximately \(53.1^\circ\). This allows for a very quick calculation once the triangle is identified. **Alternative Hack (Inscribed Angle Theorem):** 1. As established, \(AB\) is a diameter. \(O(0,0)\), \(A(8,0)\), \(B(0,6)\), \(P(8,6)\). 2. Angle \(OPA\) and angle \(OBA\) are both inscribed angles that subtend the same arc \(OA\). Therefore, \(\angle OPA = \angle OBA\). 3. Consider triangle \(OBA\). It's a right triangle at \(O\). \(OB=6\), \(OA=8\). We want \(\angle OBA\). 4. \(\tan(\angle OBA) = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{OA}{OB} = \frac{8}{6} = \frac{4}{3}\). 5. \(\angle OBA = \arctan(4/3) \approx 53.1^\circ\). Thus, \(\angle OPA = 53.1^\circ\). This is arguably the fastest method if you recall the inscribed angle theorem.