In the figure below, ECD = 25°, EAF = 30° and CFD = 35°. Find: a) CAD b) CDF c) DCF d) DFE

• Pembe $\angle ECD = 25^\circ$ inaunda arc ED. Hivyo, $arc(ED) = 2 \times \angle ECD = 2 \times 25^\circ = 50^\circ$.
• Pembe $\angle EAC = 30^\circ$ (kutoka $\angle EAF$) inaunda arc EC. Hivyo, $arc(EC) = 2 \times \angle EAC = 2 \times 30^\circ = 60^\circ$.
• Tunajua $arc(EC) = arc(ED) + arc(DC)$. $60^\circ = 50^\circ + arc(DC)$ $arc(DC) = 60^\circ - 50^\circ = 10^\circ$.
• Pembe $\angle CFD = 35^\circ$ ni pembe iliyoundwa na kodi zinazokatana AC na ED. $\angle CFD = \frac{1}{2} (arc(CD) + arc(AE))$. $35^\circ = \frac{1}{2} (10^\circ + arc(AE))$. $70^\circ = 10^\circ + arc(AE)$. $arc(AE) = 70^\circ - 10^\circ = 60^\circ$.

Kwa hivyo, vipimo vya arc ni:
$arc(CD) = 10^\circ$
$arc(DE) = 50^\circ$
$arc(EA) = 60^\circ$

a) $\angle CAD$
Pembe $\angle CAD$ inaunda arc CD.

$$\angle CAD = \frac{1}{2} arc(CD) = \frac{1}{2} \times 10^\circ = 5^\circ$$ $$\boxed{\angle CAD = 5^\circ}$$

b) $\angle CDF$
Pembe $\angle CDF$ ni sawa na $\angle CDE$. Pembe $\angle CDE$ inaunda arc CE.
$arc(CE) = arc(CD) + arc(DE) = 10^\circ + 50^\circ = 60^\circ$.

$$\angle CDE = \frac{1}{2} arc(CE) = \frac{1}{2} \times 60^\circ = 30^\circ$$ $$\boxed{\angle CDF = 30^\circ}$$

c) $\angle DCF$
Pembe $\angle DCF$ ni sawa na $\angle DCA$. Pembe $\angle DCA$ inaunda arc DA.
$arc(DA) = arc(DE) + arc(EA) = 50^\circ + 60^\circ = 110^\circ$.

$$\angle DCA = \frac{1}{2} arc(DA) = \frac{1}{2} \times 110^\circ = 55^\circ$$ $$\boxed{\angle DCF = 55^\circ}$$

d) $\angle DFE$
Pembe $\angle DFE$ na $\angle CFD$ ni pembe za ziada (ziko kwenye mstari ulionyooka).

$$\angle DFE = 180^\circ - \angle CFD = 180^\circ - 35^\circ = 145^\circ$$ $$\boxed{\angle DFE = 145^\circ}$$