Answer:
a = 130
b = 32
c = 81
d = 69
Step-by-step explanation:
Values of c and d
The diagram shows a cyclic quadrilateral, as the vertices of the quadrilateral lie on the circumference of the circle.
The opposite angles of a cyclic quadrilateral are supplementary (sum to 180°). Therefore:
[tex]\rm c^{\circ} + 99^{\circ} = 180^{\circ}\\\\c^{\circ} = 180^{\circ} - 99^{\circ}\\\\c^{\circ} = 81^{\circ}[/tex]
Therefore, the value of c is:
[tex]\Large\boxed{\boxed{\rm c = 81}}[/tex]
[tex]\rm d^{\circ} + 111^{\circ} = 180^{\circ}\\\\d^{\circ} = 180^{\circ} - 111^{\circ}\\\\d^{\circ} = 69^{\circ}[/tex]
Therefore, the value of d is:
[tex]\Large\boxed{\boxed{\rm d = 69}}[/tex]
[tex]\dotfill[/tex]
Values of a and b
An inscribed angle is the angle formed when two chords meet at one point on the circumference of a circle, and the intercepted arc is the arc that is between the endpoints of those chords.
According to the Inscribed Angle Theorem, the measure of an inscribed angle is half the measure of the intercepted arc.
For the inscribed angle 111°, the intercepted arc is (92 + a)°. Therefore, applying the inscribed angle theorem, we get:
[tex]\rm 111^{\circ}=\dfrac{1}{2}(92+a)^{\circ}[/tex]
Solve for a:
[tex]\rm 2 \cdot 111^{\circ}=2 \cdot \dfrac{1}{2}(92+a)^{\circ}\\\\\\222^{\circ}=92^{\circ}+a^{\circ}\\\\\\a^{\circ}=222^{\circ}-92^{\circ}\\\\\\a^{\circ}=130^{\circ}[/tex]
So, the value of a is:
[tex]\Large\boxed{\boxed{\rm a= 130}}[/tex]
Similarly, for the inscribed angle c°, the intercepted arc is (a + b)°. Therefore, applying the inscribed angle theorem, we get:
[tex]\rm c^{\circ}=\dfrac{1}{2}(a+b)^{\circ}[/tex]
Substitute in a = 130 and c = 81, then solve for b:
[tex]\rm 81^{\circ}=\dfrac{1}{2}(130+b)^{\circ}\\\\\\2\cdot 81^{\circ}=2\cdot\dfrac{1}{2}(130+b)^{\circ}\\\\\\162^{\circ}=130^{\circ}+b^{\circ}\\\\\\b^{\circ}=162^{\circ}-130^{\circ}\\\\\\b^{\circ}=32^{\circ}[/tex]
So, the value of b is:
[tex]\Large\boxed{\boxed{\rm b= 32}}[/tex]
