Question

In the figure, OR is perpendicular to OP and OP = 12cm. A, B and C are points on OR. If ∠OPA = 30°? ∠APB = 15°,and ∠BPC = 15°. Find OA, OB and OC. Also find AB: BC.

Answer 1

ΔOPA is a right triangle with angles 30°, 60°, 90°. Since the side opposite to 30° angle is 4√3

ie OA = \(4\sqrt{3}\) 

[\(\frac{12}{\sqrt{3}}=\frac{12\times\sqrt{3}}{\sqrt{3}\times\sqrt{3}}\) = \(\frac{12\sqrt{3}}{3}=4\sqrt{3}\) ]

Now consider ΔOPB. It is a right triangle with angles 45°, 45°, 9Q° Since the side OP = 12cm, side OB is also 12cm.

Also ΔOPC is a right triangle with angles 30°, 60°, 90°. Since the side opposite to 30° angle is 12cm, the side opposite to the 60° angle ie OC is 12√3.

Thus OA = 4√3 cm,

Trigonometry Memory Map

sin²x + cos²x = 1

sin (180 - x) = sinx

cos (180 - x) = -cosx

In a circle of radius r, the length of a chord of central angle c° = 2rsin\((\frac{c}{2})^\circ\)