Question

triangle PQR is inscribed in a circle with

centre ‘O’

, ∠Q = 70°

, ∠R = 55° and PT⟂QR.

Also, PO increased to QR meets QR at U. Find

∠TPU?​

Answer 1

Given :- A triangle PQR is inscribed in a circle with centre ‘O’, ∠Q = 70°, ∠R = 55° and PT⟂QR. Also, PO increased to QR meets QR at U.

To Find :- ∠TPU ?

Answer :-

In right angled ∆PTQ , { since PT⟂QR }

→ ∠PQT = 70° (given)

so,

→ ∠QPT = 180° - (70° + 90°) = 180° - 160° = 20° .

also, in ∆PQR,

→ ∠P = 180° - (70° + 55°) = 180° - 125° = 55°

now,

→ ∠POR = 2 * ∠PQR { Angle at centre is double of angle at circumference .}

→ ∠POR = 2 * 70° = 140° .

then, in ∆POR,

→ PO = OR { Radius.}

so,

→ ∠OPR = ∠ORP { Angle opposite to equal sides are equal in length .}

therefore,

→ ∠OPR = (180° - ∠POR)/2 = (180° - 140°)/2 = 40/2 = 20° .

hence,

→ ∠TPU = ∠QPR - (∠QPT + ∠OPR)

→ ∠TPU = 55° - (20° + 20°)

→ ∠TPU = 55° - 40°

→ ∠TPU = 15° (Ans.)

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