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

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

$R_1 = \dfrac{T_1}{\tan 14^\circ} = \dfrac{30.6}{\tan 14^\circ} = 122.73 ~ \text{m}$ ← [ 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*

## Comments

## Bakit po inadd yong 154.40 at

Bakit po inadd yong 154.40 at 6.10

## Wala akong nakita sa solution

Wala akong nakita sa solution na inadd yang dalawa. Do you mean minultiply? If yes, rectangular kasi ang shape sa protion na yan kaya minultiply para makuha ang area.

## saan po galing yung 154.40

saan po galing yung 154.40

## AB - T1.

AB-T_{1}.## yung nakuha mong 30.6 i minus

yung nakuha mong 30.6 i minus mo sa 185

## yup I noticed that too.

yup I noticed that too. siguro nagkamali lang sa solution. tayo nalang po mag adjust hehe

## Nakita ko na... doon pala sa

Nakita ko na... doon pala sa next line, na add pala talaga... yeah kayo na mag adjust... di ko na i-edit... hahaha

## san nakuha 28 degree

san nakuha 28 degree

## The azimuths are reckoned

The azimuths are reckoned from South.