Find the values of \angle BAO and AC in the second diagram.

INSIGHT

Solution:

• The diagram shows a circle with center O. AB and AC are tangents to the circle at points B and C respectively.
• We are given that AB = 7 cm and \angle BAC = 28 degrees.
• OB is the radius and is perpendicular to the tangent AB at B. So, \angle OBA = 90 degrees.
• OC is the radius and is perpendicular to the tangent AC at C. So, \angle OCA = 90 degrees.
• In triangle OBA, we have a right angle at B.
• We can find \angle BAO by bisecting \angle BAC. Since AB and AC are tangents from the same external point A, the line segment AO bisects \angle BAC.
• Therefore, \angle BAO = \angle CAO = \angle BAC / 2 = 28 degrees / 2 = 14 degrees.
• Now consider the right-angled triangle OBA. We have \angle BAO = 14 degrees and AB = 7 cm. We can find OB (radius) using the tangent function:
• \(\tan(\angle BAO) = \frac{OB}{AB}\)
• \(\tan(14^{\circ}) = \frac{OB}{7}\)
• \(OB = 7 \cdot \tan(14^{\circ})\)
• Using a calculator, \(\tan(14^{\circ}) \approx 0.2493\)
• \(OB \approx 7 \cdot 0.2493 \approx 1.745 \text{ cm}\)
• Since OC is also a radius, OC = OB \(\approx 1.745 \text{ cm}\).
• Now we need to find AC. In the right-angled triangle OCA, we have \(\\angle CAO = 14^{\circ}\) and OC \(\approx 1.745 \text{ cm}\).
• \(\tan(\angle CAO) = \frac{OC}{AC}\)
• \(\tan(14^{\circ}) = \frac{1.745}{AC}\)
• \(AC = \frac{1.745}{\tan(14^{\circ})} \approx \frac{1.745}{0.2493} \approx 7 \text{ cm}\)
• Alternatively, since \(\angle BAO = \angle CAO\) and \(\angle OBA = \angle OCA = 90^{\circ}\), triangle OBA is congruent to triangle OCA (by Angle-Side-Angle, or more precisely, Angle-Angle-Side if we consider the shared hypotenuse AO and equal angles \(\angle BAO = \angle CAO\)). Therefore, AB = AC = 7 cm.

Final Answer:

\angle BAO = 14^{\circ}

AC = 7 \text{ cm}