In the figure below PQ is the diameter of the circle with centre O. Calculate the sizes of the indicated angles.

Haya, tutasuluhisha maswali haya hatua kwa hatua.

3. Tafuta ukubwa wa pembe $a$, $b$, na $c$.

• Hatua ya 1: Tafuta angle $a$. $PQ$ ni kipenyo cha duara. Pembe inayotegemea arc $QR$ kwenye mzingo ni $\angle QPR$ na $\angle QSR$. Pembe zinazotegemea arc sawa kwenye mzingo ni sawa. Kwa hiyo, $a = \angle QSR = \angle QPR = 50^\circ$.

• Hatua ya 2: Tafuta angle $b$. $O$ ni kituo cha duara. $OQ = OS$ (radii). Hivyo, $\triangle OQS$ ni pembetatu pacha. Pembe kwenye kituo $\angle QOS$ ni mara mbili ya pembe kwenye mzingo $\angle QPS$ inayotegemea arc sawa $QS$. Kwanza, tafuta $\angle QPS$. Kwa kuwa $PQ$ ni kipenyo, $\angle PSQ = 90^\circ$ (pembe katika nusu duara). Katika $\triangle PQS$, jumla ya pembe ni $180^\circ$.

  $$\angle QPS + \angle PQS + \angle PSQ = 180^\circ$$ $$\angle QPS + 35^\circ + 90^\circ = 180^\circ$$ $$\angle QPS + 125^\circ = 180^\circ$$ $$\angle QPS = 180^\circ - 125^\circ = 55^\circ$$

  Sasa, tafuta $b = \angle QOS$:

  $$b = 2 \times \angle QPS$$ $$b = 2 \times 55^\circ$$ $$b = 110^\circ$$

• Hatua ya 3: Tafuta angle $c$. Pembe kwenye kituo $\angle QOR$ ni mara mbili ya pembe kwenye mzingo $\angle QPR$ inayotegemea arc sawa $QR$.

  $$\angle QOR = 2 \times \angle QPR$$ $$\angle QOR = 2 \times 50^\circ$$ $$\angle QOR = 100^\circ$$

  $PQ$ ni kipenyo, hivyo $\angle POQ = 180^\circ$. Tunaweza kuandika $\angle POQ$ kama jumla ya pembe $\angle POS$, $\angle ROS$, na $\angle ROQ$. Au, $\angle POQ = \angle POS + \angle SOQ$. Tunajua $\angle QOS = b = 110^\circ$. Tunajua $\angle QOR = 100^\circ$. Kutoka mchoro, $S$ iko kati ya $P$ na $R$ kwenye arc. Hivyo, $\angle ROS = \angle QOS - \angle QOR$. (Hii inategemea mpangilio wa pointi. Ikiwa $R$ iko kati ya $P$ na $S$, basi $\angle ROS = \angle QOS - \angle QOR$ haitafanya kazi). Hebu tutumie pembe kwenye mzingo. Arc $RS$ inategemea $\angle RPS$ na $\angle RQS$. $\angle RPS = \angle QPS - \angle QPR = 55^\circ - 50^\circ = 5^\circ$. Kwa hiyo, pembe kwenye kituo $\angle ROS$ ni mara mbili ya $\angle RPS$.

  $$c = \angle ROS = 2 \times \angle RPS$$ $$c = 2 \times 5^\circ$$ $$c = 10^\circ$$

  Au, $\angle RQS = \angle PQR - \angle PQS$. Katika $\triangle PQR$, $\angle PRQ = 90^\circ$. $\angle PQR = 180^\circ - 90^\circ - 50^\circ = 40^\circ$. $\angle RQS = 40^\circ - 35^\circ = 5^\circ$. $c = \angle ROS = 2 \times \angle RQS = 2 \times 5^\circ = 10^\circ$.

  Pembe $a$ ni $\boxed{\text{50}^\circ}$, pembe $b$ ni $\boxed{\text{110}^\circ}$, na pembe $c$ ni $\boxed{\text{10}^\circ}$.

4. Tafuta angle $\angle CAB$.

• Hatua ya 1: Tumia pembetatu pacha $\triangle OAB$. $O$ ni kituo cha duara, hivyo $OA = OB$ (radii). Hivyo, $\triangle OAB$ ni pembetatu pacha. $\angle OBA = \angle BAO$. Tunapewa $\angle BAO = 30^\circ$. Kwa hiyo, $\angle OBA = 30^\circ$.

• Hatua ya 2: Tafuta $\angle OBC$. Tunapewa $\angle ABC = 40^\circ$. $\angle OBC = \angle ABC - \angle OBA$ $\angle OBC = 40^\circ - 30^\circ$ $\angle OBC = 10^\circ$.

• Hatua ya 3: Tumia pembetatu pacha $\triangle OBC$. $OB = OC$ (radii). Hivyo, $\triangle OBC$ ni pembetatu pacha. $\angle OCB = \angle OBC = 10^\circ$.

• Hatua ya 4: Tumia pembetatu pacha $\triangle OAC$. $OA = OC$ (radii). Hivyo, $\triangle OAC$ ni pembetatu pacha. $\angle OAC = \angle OCA$. Wacha $\angle OAC = x$.

• Hatua ya 5: Tumia jumla ya pembe katika $\triangle ABC$. $\angle CAB = \angle BAO + \angle OAC = 30^\circ + x$. $\angle BCA = \angle BCO + \angle OCA = 10^\circ + x$. Jumla ya pembe katika $\triangle ABC$ ni $180^\circ$.

  $$\angle CAB + \angle ABC + \angle BCA = 180^\circ$$ $$(30^\circ + x) + 40^\circ + (10^\circ + x) = 180^\circ$$ $$80^\circ + 2x = 180^\circ$$ $$2x = 180^\circ - 80^\circ$$ $$2x = 100^\circ$$ $$x = 50^\circ$$

• Hatua ya 6: Tafuta $\angle CAB$. $\angle CAB = 30^\circ + x = 30^\circ + 50^\circ = 80^\circ$. Angle $\angle CAB$ ni $\boxed{\text{80}^\circ}$.

5. Tafuta angle $\angle QOP$.

• Hatua ya 1: Tumia mistari sambamba. Tunapewa $QP \parallel RO$. $OQ$ ni mstari unaokatisha mistari sambamba. Kwa hiyo, pembe mbadala za ndani ni sawa. $\angle PQO = \angle QOR$. Tunapewa $\angle QOR = 80^\circ$. Kwa hiyo, $\angle PQO = 80^\circ$.

• Hatua ya 2: Tumia pembetatu pacha $\triangle OQP$. $O$ ni kituo cha duara, hivyo $OQ = OP$ (radii). Hivyo, $\triangle OQP$ ni pembetatu pacha. Pembe za msingi za pembetatu pacha ni sawa. $\angle OPQ = \angle PQO$. Kwa kuwa $\angle PQO = 80^\circ$, basi $\angle OPQ = 80^\circ$.

• Hatua ya 3: Tafuta $\angle QOP$. Jumla ya pembe katika pembetatu $\triangle OQP$ ni $180^\circ$.

  $$\angle QOP + \angle OPQ + \angle PQO = 180^\circ$$ $$\angle QOP + 80^\circ + 80^\circ = 180^\circ$$ $$\angle QOP + 160^\circ = 180^\circ$$ $$\angle QOP = 180^\circ - 160^\circ$$ $$\angle QOP = 20^\circ$$

  Angle $\angle QOP$ ni $\boxed{\text{20}^\circ}$.