7) In the given circle, the angle BOC is 68 degrees. Find x and y.

Solution:

• The angle subtended by an arc at the center is double the angle subtended by it at any point on the remaining part of the circle.
• The angle subtended by arc BC at the center is $$\angle BOC = 68^{\circ}$$.
• Therefore, the angle subtended by arc BC at point D on the circumference is $$\angle BDC = \frac{1}{2} \angle BOC = \frac{1}{2} \times 68^{\circ} = 34^{\circ}$$.
• From the figure, $$y = \angle BDC$$, so $$y = 34^{\circ}$$.
• Also, $$\angle BOC$$ is a straight angle. Therefore, $$\angle BOC = 180^{\circ}$$ if it is a diameter. However, O is the center of the circle and BC is a chord.
• The angle $$x$$ is $$\angle BAC$$. It subtends the arc BC. Thus, $$x = \angle BDC = 34^{\circ}$$.
• Alternatively, $$\angle BAC$$ and $$\angle BDC$$ subtend the same arc BC. Thus, $$x = y$$.
• The angle AOC is a straight line passing through the center O. This means AC is a diameter.
• If AC is a diameter, then $$\angle ABC = 90^{\circ}$$ and $$\angle ADC = 90^{\circ}$$.
• In triangle BOC, OB = OC (radii), so it is an isosceles triangle. $$\angle OBC = \angle OCB = \frac{180^{\circ} - 68^{\circ}}{2} = \frac{112^{\circ}}{2} = 56^{\circ}$$.
• In triangle AOC, OA = OC (radii), so it is an isosceles triangle.
• The angle $$\angle ABC$$ is an inscribed angle subtending arc AC. If AC is a diameter, then $$\angle ABC = 90^{\circ}$$.
• In $$\triangle BOC$$, OB=OC, so $$\angle OBC = \angle OCB = (180-68)/2 = 56^{\circ}$$.
• In $$\triangle AOD$$, OA=OD, so $$\angle OAD = \angle ODA$$.
• The angle subtended by arc CD at the center is $$\angle COD$$.
• The angle subtended by arc AB at the center is $$\angle AOB$$.
• We are given $$\angle BOC = 68^{\circ}$$. $$x = \angle BAC$$ and $$y = \angle BDC$$. Both subtend arc BC. So $$x=y$$.
• Since AC is a diameter, $$\angle ADC = 90^{\circ}$$. So $$y=90^{\circ}$$ is incorrect based on the visual.
• Let's assume AC is a diameter. Then $$\angle ABC = 90^{\circ}$$.
• The angle subtended by arc BC at the center is $$68^{\circ}$$. Therefore, the angle subtended at the circumference by arc BC is $$x = y = 68^{\circ}/2 = 34^{\circ}$$.

Ответ: $$x = 34^{\circ}$$, $$y = 34^{\circ}$$