Question

answer plzz today maths board exam paper cbse​

Answer 1

It is from tb ncert example 3 problem

It is a direct question

Answer 2

Use Triangle property:

As you know, OP = radius = 5 cm

PQ = 8 cm

PR = RQ = 4 cm

Using Pythagoras:

(PO)² = (OR)² + (PR)²

OR = √(5² - 4²) = 3 cm

PT = TQ

In triangle ∆OPR,

Let angle POR be "ø"

So, cos ø = (OR/OP)

Also, angle POT = ø

In ∆TOP,

cos(angle POT) = cos ø = PO/OT

Using Pythagoras in ∆ TOP,

(OT)² = (PT)² + (OP)²

PT = √((OP)² - (OP)²)

Hence, find "PT".

Thankyou!!!