In the given figure, PC is a tangent to the circle at C. AOB is the diameter which when extended meets the ta…

Question

In the given figure, PC is a tangent to the circle at C. AOB is the diameter which when extended meets the tangent at P. Find $\angle BCO$ and $\angle CBA$, if $\angle PCA = 110^\circ$.

Accepted Answer

## Model Answer **Given:** ∠PCA = 110°, PC is tangent at C, AOB is diameter. **Step 1:** OC ⊥ PC (radius ⊥ tangent) ∴ ∠OCA = 90° **Step 2:** ∠PCO = ∠PCA − ∠OCA = 110° − 90° = **20°** ∴ **∠BCO = ∠PCO = 20°** ... (OC = OB = radius, so ∠OCB = ∠OBC; but since ∠PCO = 20°, and OC = OB, △OCB is isosceles) Wait — ∠BCO = ∠OCA − ∠OCA... Let me use: ∠ACO = 90°, so ∠ACB = 90° (angle in semicircle). **∠BCO:** ∠OCB = ∠OBC (OC = OB = radii, △OCB isosceles) ∠PCA = 110° ⟹ ∠ACO = 90° − ∠PCO... **Correct working:** ∠OCP = 90° (radius ⊥ tangent) ∠OCA = ∠OCP − ∠PCA ... no, ∠PCA = 110° > 90° So: ∠OCA = ∠PCA − ∠PCO ∠PCO = 90° ⟹ ∠OCA = 110° − 90° = **20°** Since OC = OA (radii), △OCA is isosceles ⟹ ∠OAC = ∠OCA = 20° ∠AOC = 180° − 20° − 20° = 140° ∠BOC = 180° − 140° = 40° (AOB is straight line) Since OC = OB (radii), △OCB is isosceles: ∠**BCO** = ∠OBC = (180° − 40°)/2 = **70°** **∠CBA:** ∠CBA = ∠OBC = **70°** *(Or: ∠ACB = 90°, angle in semicircle; ∠CAB = 20°; ∴ ∠CBA = 70°)* **Answers: ∠BCO = 70°, ∠CBA = 70°**…

Mathematics — Standard (041) — CBSE Class 10 board question