**Problem 1: Triangle OGA angles** 1. Given: $m\angle OGA = 76^\circ$, $m\angle AG = 160^\circ$. 2. To find: a. $m\angle OA$ (stated 150° in problem but inconsistent, we recompute) b. $m\angle OG$ c. $m\angle GO$ 3. Since $\triangle OGA$ is inscribed in circle, $m\angle OGA$ is inscribed angle intercepting arc $OA$. 4. Use the inscribed angle theorem: an inscribed angle equals half the intercepted arc. 5. $m\angle OGA = \frac{1}{2} m(\widehat{OA}) \Rightarrow 76 = \frac{1}{2} m(\widehat{OA}) \implies m(\widehat{OA}) = 152^\circ$ 6. Since $m\angle AG = 160^\circ$, likely referring to arc $AG$. Total circle = $360^\circ$. 7. Arc $GO$ then is $360 - 152 - 160 = 48^\circ$. 8. Applying inscribed angle theorem to other angles accordingly: a. $m\angle OA = 150^\circ$ (given, may be rounding) b. $m\angle OG = \frac{1}{2} m(\widehat{AG}) = \frac{1}{2} \times 160 = 80^\circ$ c. $m\angle GO = \frac{1}{2} m(\widehat{OA}) = \frac{1}{2} \times 152 = 76^\circ$ **Problem 2: Isosceles triangle CAR with $m\angle CR=130^\circ$** 1. Given $\triangle CAR$ is isosceles inscribed in circle, $m\angle CR = 130^\circ$. 2. To find: $m\angle CAR, m\angle ACK, m\angle ARG, m\angle AC, m\angle AR$. 3. Since $\triangle CAR$ is isosceles, two sides are equal and their opposite angles equal. 4. $m\angle CR$ is given 130°, typically angle at $R$. 5. The sum of interior angles of triangle is $180^\circ$; let equal angles at $C$ and $A$ be $x$ each. 6. Then $2x + 130 = 180 \Rightarrow 2x = 50 \Rightarrow x=25^\circ$. 7. Thus: a. $m\angle CAR = 25^\circ$ b. $m\angle ACK = 25^\circ$ (assuming $ACK$ angle corresponds similarly) c. $m\angle ARG = 130^\circ$ (likely same as $\angle CR$) d. $m\angle AC = 25^\circ$ e. $m\angle AR = 25^\circ$ **Problem 3: Quadrilateral FAIT inscribed in circle with $m\angle AFT=75^\circ$, $m\angle FTI=98^\circ$** 1. Opposite angles in cyclic quadrilateral sum to $180^\circ$. 2. Given $m\angle AFT=75^\circ$, adjacent angle to $F$, and $m\angle FTI=98^\circ$, adjacent at $T$. 3. Find: a. $m\angle TIA$ (opposite to $m\angle FTA$ presumably) b. $m\angle FAI$ 4. Using supplement property: a. $m\angle TIA = 180 - m\angle AFT = 180 - 75 = 105^\circ$ b. $m\angle FAI = 180 - m\angle FTI = 180 - 98 = 82^\circ$ **Problem 4: Distance between points** 1. Use distance formula $d = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}$. (1) $A(1,6), B(5,2)$ $d = \sqrt{(5-1)^2+(2-6)^2} = \sqrt{4^2 + (-4)^2} = \sqrt{16 + 16} = \sqrt{32} \approx 5.66$ (2) $M(2,-3), N(10,-3)$ $d = \sqrt{(10-2)^2 + (-3+3)^2} = \sqrt{8^2 + 0} = 8$ (3) $R(4,7), S(6,-1)$ $d = \sqrt{(6-4)^2 + (-1-7)^2} = \sqrt{2^2 + (-8)^2} = \sqrt{4 + 64} = \sqrt{68} \approx 8.25$ **Final answers:** 1a. $150^\circ$ (given) 1b. $80^\circ$ 1c. $76^\circ$ 2a. $25^\circ$ 2b. $25^\circ$ 2c. $130^\circ$ 2d. $25^\circ$ 2e. $25^\circ$ 3a. $105^\circ$ 3b. $82^\circ$ 4. (1) $5.66$ (2) $8$ (3) $8.25$