Final answer:
In a triangle circumscribed around a circle, the lengths of the sides of the triangle are equal to the tangent of half the measure of the intercepted arcs. By using this property, the values of CQ, AB, and RT can be calculated.
Step-by-step explanation:
In a triangle circumscribed around a circle, the lengths of the sides of the triangle are equal to the tangent of half the measure of the intercepted arcs. Using this property, we can solve for CQ, AB, and RT.
Given that CB = 9, NT = 7, and RQ = 10, we can solve for CQ:
Since CQ = CN + NQ, we can use the tangent of half the intercepted arcs to find CN and NQ. In this case, the intercepted arcs are CB and NT.
Taking half the measure of CB and NT, we get θ = 1/2(TAN⁻¹(9/2)) + 1/2(TAN⁻¹(7/2)). Using the tangent addition formula, TAN(θ) = (TAN⁻¹(9/2) + TAN⁻¹(7/2)) / (1 - (TAN⁻¹(9/2) * TAN⁻¹(7/2))), we can find the value of TAN(θ).
Once we have the value of TAN(θ), we can solve for CN and NQ using the formula CN = CQ * TAN(θ) and NQ = CQ * TAN(θ).
Therefore, CQ = CN + NQ = CQ * TAN(θ) + CQ * TAN(θ). We can solve for CQ using algebraic manipulation.
Similarly, we can use the tangent of half the intercepted arcs to find AB and RT. The intercepted arcs in this case are CB and RQ.
Therefore, AB = CB * TAN(θ) + CB * TAN(θ) and RT = RQ * TAN(θ) + RQ * TAN(θ).
Now we can substitute the given values into the equations to find the values of CQ, AB, and RT.