Mathematics 10 — Exercise Review · Q3, d

Full question

SQ is a diameter and QSR=72°. Find d=SQR.

OSQPR∠PSQ=42°; ∠QSR=72°a=SPQ; b=SRQ; c=PQS; d=SQRDiagram not to scale

Step-by-step solution

  1. A radius is perpendicular to the tangent at its point of contact, so the contact angle is 90°.

  2. The three angles of a triangle add to 180°. Subtract both known angles.

  3. Calculate this operation before combining the remaining terms.

    (-1)×90=-90
  4. Calculate this operation before combining the remaining terms.

    (-1)×72=-72
  5. Substitute the given numbers into the formula.

    180−90−72
  6. Calculate this operation before combining the remaining terms.

    180−90−72=18
  7. The required result is shown below. Unit: °

    18

Answer

The final answer is shown below.

18