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Ex 11.2, 4 (Concept)

Draw a pair of tangents to a circle of radius 5 cm which are inclined to each other at an angle of 60°.

Concept

Given angle between tangents is 60°

i.e. ∠ QPR = 60°

⇒ ∠ OQR = 2 × 60° = 120°

So, we need to draw ∠ OQR = 120°

Also,

OQ ⊥ QP & OR ⊥ PR

Thus, to make tangents, we draw perpendiculars from point Q and R

 

Ex 11.2, 4

Draw a pair of tangents to a circle of radius 5 cm which are inclined to each other at an angle of 60°.

The tangents can be constructed in the following manner:

  1. Draw a circle of radius 5 cm with centre O.
  2. We need make angle of 120° at center
  3. Draw diameter QOA
  4. Draw ∠ OAR = 60°
  5. Thus, ∠ QOR = 180° – ∠ OAR = 180 – 60 = 120°
  6. Therefore, our two tangents will touch the circle at Q and R
  7. Now, we know that tangent is perpendicular to radius. Draw perpendicular from point Q and point R
  8. Let P be point where both perpendiculars  intersect

Thus, PQ and PR are the required tangents at angle of 60°

Justification

We need to prove PQ and PR are tangents and ∠ QPR = 60°

Since PQ is perpendicular to OQ (radius)

PQ must be a tangent

Also, PR is a tangent

Now, we need to prove ∠ QPR = 60°

Since PQ ⊥ OQ, ∠ OQP = 90°

Also, PR ⊥ OR,  ∠ ORP = 90°

and ∠ QOR = 120°

Now, in quadrilateral OQPR

∠ QPR + ∠ PQO + ∠ PRO + ∠ QOR = 360°

∠ QPR + 90° + 90° + 120°  = 360°

∠ QPR + 300°  = 360°

∠ QPR = 360° – 300°

∠ QPR = 60°

Thus, angle between tangents is 60°

Now, in quadrilateral OQPR

∠ QPR + ∠ PQO + ∠ PRO + ∠ QOR = 360°

∠ QPR + 90° + 90° + 120°  = 360°

∠ QPR + 300°  = 360°

∠ QPR = 360° – 300°

∠ QPR = 60°

Thus, angle between tangents is 60°

-ev-

  1. Chapter 11 Class 10 Constructions
  2. Serial order wise
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CA Maninder Singh
CA Maninder Singh is a Chartered Accountant for the past 7 years. He provides courses for Practical Accounts, Taxation and Efiling at teachoo.com .
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