Ap, pa pb to a0. if op is to of ,is .

have a nice time reading and understanding his:)

Step-by-step explanation:

considering APB as the triangle

AP is the tangent to the circle.

∴ OA ⊥ AP  (Radius is perpendicular to the tangent at the point of contact)

⇒ ∠OAP = 90º  

In Δ OAP,

sin ∠OPA = OA/OP = r/2r [Diameter of the circle]

∴ sin ∠OPA = 1/2 = sin 30º

⇒ ∠OPA =  30º

Similarly, it can he prayed that ∠OPB = 30

How, LAB = LOP + LOB = 30° + 30° = 60°  

In APB,  

PA = PB [lengths &tangents drawn from an external point to circle areequal]  

⇒ ∠PAB = ∠PBA --- (1) [Equal sides have equal angles apposite to them]  

∠PAB + ∠PBA + ∠APB = 180° [Angle sum property]  

∠PAB + ∠PBA + ∠APB = 180° - 60° [Using (1)]  

⇒ 2∠PAB = 120°

⇒ ∠PAB = 60°  

From (1) and (2)  

∠PAB = ∠PBA = ∠APB = 60°

APB is an equilateral triangle.

i can answer this question sweetie!

step-by-step explanation:

because linear equations are a lot simpler than quadratics equations, and linear equations formula is a lot easier to memorize. that's why i think they're harder.