|
(a)
Given the area \(\Delta{BCD}=12\) cm\(^2\).
\(\begin{aligned} \dfrac{1}{2}(BC)(CD)\sin{C}&=12 \\ \dfrac{1}{2}(7)(4)\sin{C}&=12 \\ 14\sin{C}&=12 \\ \sin{C}&=\dfrac{12}{14} =0.8571 \\\\C&=59^\circ. \end{aligned}\)
\(\therefore \angle{BCD}=59^\circ.\)
(b)
Apply cosine rule,
\(\begin{aligned} BD^2&=CD^2+BC^2-2(4)(7)\cos{59^\circ} \\ BD^2&=4^2+7^2-2(4)(7)\cos{59^\circ}\\ BD^2&=65-28.84\\ BD^2&=36.16 \end{aligned}\)
\(\therefore BD=6.013\text{ cm}.\)
(c)
Apply sine rule,
\(\begin{aligned} \dfrac{AB}{\sin{35^\circ}}&=\dfrac{6.013}{\sin{A}}\\ \dfrac{10}{\sin{35^\circ}}&=\dfrac{6.013}{\sin{A}}\\ \sin{A}&=\dfrac{6.013\times\sin{35^\circ}}{10}\\ \sin{A}&=0.3449\\\\ A&=20.18^\circ. \end{aligned}\)
\(\angle{ABD}=180^\circ-35^\circ-20.18^\circ\)
\(\therefore \angle{ABD}=124.82^\circ.\)
(d)
Area of quadrilateral \(ABCD\),
\(\begin{aligned} &=\text{Area }\Delta{ABD}+\text{Area }\Delta{BCD} \\ &=\dfrac{1}{2}(AB)(BD)\sin{B}+12\text{ cm}^2 \\ &=\dfrac{1}{2}(10)(6.013)\sin{124.82^\circ}+12 \\ &=24.68+12 \\ &=36.68\text{ cm}^2. \end{aligned}\)
|
