Question

PAQ is a tangent to the circle with centre o at a point A as shown in figure. if LOBA=35digree find the value of LBAQ and LACB​

Answer 1

Step-by-step explanation:

In ∆OAB ∵ OA = OB (radii of a circle) ∴ ∠OBA = ∠OAB (Opposite angle of the equal side of triangle) ∴ ∠OAB = 32° ⇒ ∠x = 32° (∠OBA = 32°) …..(i) Again, PAQ the tangent at point A of circle and OA is radius. ∴ OA ⊥ PQ ⇒ ∠OAQ = 90° ∴ ∠BAQ + ∠OAB = 90° ∴ ∠BAQ + 32° = 90° [∵ ∠OAB = 32°] ∠BAQ = 90° – 32° = 58° ∠BAQ = ∠ACB [angle made in alternate segment] 58° = ∠y ∠x = 32°, ∠y = 58°Read more on Sarthaks.com - https://www.sarthaks.com/762568/in-the-following-figure-o-is-the-center-of-circle-paq-is-the-tangent-of-circle-at-point-a?show=762569#a762569