1. Problem 1: Given point $P(1,4)$, a direction angle from $P$ to $Q$ of $128^\circ43'$ and distance $PQ=5$ cm, find the coordinates of point $Q$. Step 1: Convert angle $128^\circ43'$ to decimal degrees. $$128^\circ43' = 128 + \frac{43}{60} = 128.7167^\circ$$ Step 2: Calculate the change in coordinates using $$\Delta x = r \cos \theta = 5 \cos 128.7167^\circ$$ $$\Delta y = r \sin \theta = 5 \sin 128.7167^\circ$$ Step 3: Evaluate these values (using degrees): $$\cos 128.7167^\circ \approx -0.6225$$ $$\sin 128.7167^\circ \approx 0.7826$$ So, $$\Delta x = 5 \times (-0.6225) = -3.1125$$ $$\Delta y = 5 \times 0.7826 = 3.9130$$ Step 4: Find coordinates of $Q$: $$x_Q = x_P + \Delta x = 1 - 3.1125 = -2.1125$$ $$y_Q = y_P + \Delta y = 4 + 3.9130 = 7.9130$$ Final answer for Q: $$Q(-2.11,\ 7.91)$$ 2. Problem 2: Given a triangle with vertices and measurements: $AD = 1.75$ m, $BD=4$ m, $DC=4$ m, $E$ and $F$ each $0.5$ m. Find (a) angle $\alpha$, (b) length $AB$, (c) length $EB$, (d) total length $AE$. Step 1: Calculate angle $\alpha$ in right triangle $ABD$. Using tangent, $$\tan \alpha = \frac{AD}{BD} = \frac{1.75}{4} = 0.4375$$ $$\alpha = \arctan 0.4375 \approx 23.62^\circ$$ Step 2: Calculate length $AB$ using Pythagoras: $$AB = \sqrt{AD^2 + BD^2} = \sqrt{1.75^2 + 4^2} = \sqrt{3.0625 + 16} = \sqrt{19.0625} \approx 4.37 \text{ m}$$ Step 3: Calculate length $EB$: Since points $E$ and $F$ are each $0.5$ m horizontally from $B$ and $C$, and $BD=4$ m, $$EB = BD - 0.5 = 4 - 0.5 = 3.5 \text{ m}$$ Step 4: Calculate total length $AE$ as sum of $AB$ and $BE$: $$AE = AB + EB = 4.37 + 3.5 = 7.87 \text{ m}$$ 3. Problem 3: Road cross slope is 2% on a 13 m wide road. Find (a) slope angle in degrees, (b) height $t$ at center. Step 1: Convert slope percentage to ratio: $$\text{slope} = 2\% = 0.02$$ Step 2: Find angle using tangent (small angle approximation): $$\theta = \arctan 0.02 \approx 1.15^\circ$$ Step 3: Calculate height $t$ at center: Slope means height difference $t$ over half the road width (since slope is on both sides): $$t = 0.02 \times \frac{13}{2} = 0.02 \times 6.5 = 0.13 \text{ m}$$ Final answers: 1. $Q(-2.11,\ 7.91)$ 2a. $\alpha = 23.62^\circ$ 2b. $AB = 4.37$ m 2c. $EB = 3.5$ m 2d. $AE = 7.87$ m 3a. $\theta = 1.15^\circ$ 3b. $t=0.13$ m