ABCD is a trapezium which AB is parallel to DC. DC = 64 cm, angle BAD = 45 and angle BDC = 38. Calculate correct to 1 decimal place: a) The length of CB. b) The length of AB. c) The perpendicular distance of D from AB. d) The radius of a circle that passes through the point BC and D.

$$CB^2 = BD^2 + DC^2 - 2 \cdot BD \cdot DC \cdot \cos(\angle BDC)$$ $$CB^2 = (10.9)^2 + (64)^2 - 2 \cdot (10.9) \cdot (64) \cdot \cos(38^\circ)$$ $$CB^2 = 118.81 + 4096 - 1395.2 \cdot (0.78801)$$ $$CB^2 = 4214.81 - 1099.50766$$ $$CB^2 = 3115.30234$$

$$CB = \sqrt{3115.30234}$$ $$CB \approx 55.8149 \text{ cm}$$

Ikiwa imerekebishwa kwa sehemu moja ya desimali:

$$CB \approx \boxed{55.8 \text{ cm}}$$

$\angle ABD = \angle BDC = 38^\circ$.
Katika pembetatu ABD, jumla ya pembe ni $180^\circ$.

$$\angle ADB = 180^\circ - \angle BAD - \angle ABD$$ $$\angle ADB = 180^\circ - 45^\circ - 38^\circ$$ $$\angle ADB = 97^\circ$$

$$\frac{AB}{\sin(\angle ADB)} = \frac{BD}{\sin(\angle BAD)}$$ $$\frac{AB}{\sin(97^\circ)} = \frac{10.9}{\sin(45^\circ)}$$ $$AB = \frac{10.9 \cdot \sin(97^\circ)}{\sin(45^\circ)}$$ $$AB = \frac{10.9 \cdot 0.99255}{0.70711}$$ $$AB = \frac{10.818795}{0.70711}$$ $$AB \approx 15.2999 \text{ cm}$$

Ikiwa imerekebishwa kwa sehemu moja ya desimali:

$$AB \approx \boxed{15.3 \text{ cm}}$$

$$\frac{AD}{\sin(\angle ABD)} = \frac{BD}{\sin(\angle BAD)}$$ $$\frac{AD}{\sin(38^\circ)} = \frac{10.9}{\sin(45^\circ)}$$ $$AD = \frac{10.9 \cdot \sin(38^\circ)}{\sin(45^\circ)}$$ $$AD = \frac{10.9 \cdot 0.61566}{0.70711}$$ $$AD = \frac{6.709794}{0.70711}$$ $$AD \approx 9.4889 \text{ cm}$$

Umbali wa wima $h$ kutoka D hadi AB ni $AD \cdot \sin(\angle BAD)$.

$$h = AD \cdot \sin(45^\circ)$$ $$h = 9.4889 \cdot 0.70711$$ $$h \approx 6.7099 \text{ cm}$$

Ikiwa imerekebishwa kwa sehemu moja ya desimali:

$$h \approx \boxed{6.7 \text{ cm}}$$

Tunajua $BD = 10.9$ cm, $DC = 64$ cm, $CB \approx 55.8149$ cm, na $\angle BDC = 38^\circ$.
Tumia sheria ya Saini kupata $\angle CBD$:

$$\frac{DC}{\sin(\angle CBD)} = \frac{CB}{\sin(\angle BDC)}$$ $$\sin(\angle CBD) = \frac{DC \cdot \sin(\angle BDC)}{CB}$$ $$\sin(\angle CBD) = \frac{64 \cdot \sin(38^\circ)}{55.8149}$$ $$\sin(\angle CBD) = \frac{64 \cdot 0.61566}{55.8149}$$ $$\sin(\angle CBD) = \frac{39.40224}{55.8149} \approx 0.70593$$

Kuna uwezekano wa pembe mbili: $\angle CBD_1 = \arcsin(0.70593) \approx 44.88^\circ$ au $\angle CBD_2 = 180^\circ - 44.88^\circ = 135.12^\circ$.
Ili kuhakikisha, tunachunguza jumla ya pembe katika pembetatu. Ikiwa $\angle CBD = 44.88^\circ$, basi $\angle BCD = 180^\circ - 38^\circ - 44.88^\circ = 97.12^\circ$. Hii inamaanisha upande mfupi zaidi $BD$ (10.9 cm) ungekuwa kinyume na pembe kubwa zaidi ($97.12^\circ$), ambayo si sahihi.
Kwa hivyo, $\angle CBD = 135.12^\circ$.
Sasa, $\angle BCD = 180^\circ - 38^\circ - 135.12^\circ = 6.88^\circ$.

$$R = \frac{CB}{2 \sin(\angle BDC)}$$ $$R = \frac{55.8149}{2 \cdot \sin(38^\circ)}$$ $$R = \frac{55.8149}{2 \cdot 0.61566}$$ $$R = \frac{55.8149}{1.23132}$$ $$R \approx 45.328 \text{ cm}$$

Ikiwa imerekebishwa kwa sehemu moja ya desimali:

$$R \approx \boxed{45.3 \text{ cm}}$$