Reversed Curve to Connect Three Traversed Lines

$T_1 = \frac{1}{4}BC = \frac{1}{4}(122.4) = 30.6 ~ \text{m}$

$T_2 = 122.4 - 30.6 = 91.8 ~ \text{m}$

$R_1 = \dfrac{T_1}{\tan 14^\circ} = \dfrac{30.6}{\tan 14^\circ} = 122.73 ~ \text{m}$   ←   [ A ]   answer for part 1

$R_2 = \dfrac{T_2}{\tan 30^\circ} = \dfrac{91.8}{\tan 30^\circ} = 159.00 ~ \text{m}$
 

 

Total length of road from A to D
$L = 154.40 + \dfrac{2\pi(122.73)(28^\circ)}{180^\circ} + \dfrac{\pi(159)(60^\circ)}{180^\circ} + 193.20$

$L = 154.40 + 59.98 + 166.50 + 193.20$

$L = 574.08 ~ \text{m}$   ←   [ C ]   answer for part 2
 

 

Area of road pavement
$\begin{align}
A = & 154.40(6.10) + \dfrac{\pi \left[ (119.68 + 6.10)^2 - 119.68^2 \right](28^\circ)}{360^\circ} \\
& + \dfrac{\pi \left[ (155.95 + 6.10)^2 - 155.95^2 \right](60^\circ)}{360^\circ} + 193.20(6.10)
\end{align}$

$A = 160.50 + 365.86 + 1015.68 + 1178.52$

$A = 2720.56 ~ \text{m}^2$
 

Cost of concrete pavement
$C = 1,800(2720.56) = \text{P}4,897,008$   ←   [ D ]   answer for part 3