1. Stating Question 2: A compound curve has a common tangent length of 81.6 m, making angles $15^\circ$ and $17^\circ$ with the tangents of the first and second curves respectively. The tangent length of the second curve is 41.2 m. Find the radius of the first curve. 2. For the first curve, let radius be $R_1$, and for the second curve radius $R_2$. 3. The length of tangent $T$ to a curve radius $R$ and central angle $\Delta$ is $T = R \tan \frac{\Delta}{2}$. 4. Since $T_2 = 41.2 m$ and angle with the second curve is $17^\circ$, we relate this to radius $R_2$, but we only need $R_1$ so focus on the geometry: The common tangent length $L = 81.6 m$ and angles with the first and second curves are $15^\circ$ and $17^\circ$. 5. Using the cosine rule in the triangle formed by the common tangent and the tangents to curves: $$L = R_1 \tan 15^\circ + R_2 \tan 17^\circ$$ Also given $T_2 = R_2 \tan 17^\circ = 41.2$ Solve for $R_2$: $$R_2 = \frac{41.2}{\tan 17^\circ}$$ Calculate $\tan 17^\circ \approx 0.3057$, so: $$R_2 \approx \frac{41.2}{0.3057} = 134.8\,m$$ 6. Substitute $R_2$ back in the first equation: $$81.6 = R_1 \tan 15^\circ + 41.2$$ Calculate $\tan 15^\circ \approx 0.2679$: $$81.6 - 41.2 = R_1 \times 0.2679$$ $$40.4 = 0.2679 R_1$$ Solve for $R_1$: $$R_1 = \frac{40.4}{0.2679} \approx 150.8\,m$$ --- 7. Stating Question 3: A compound curve's common tangent makes angles $12^\circ$ and $18^\circ$ with the tangents of the first and second curves respectively. The chord lengths PC to PCC = 71.8 m, PCC to PT = 49.5 m. Find the chord length PC to PT parallel to the common tangent. 8. Let the chord $PC-PT$ make angle $\theta$, the sum of internal angles with chords $PC-PCC$ and $PCC-PT$. 9. Since $PC-PCC$ and $PCC-PT$ chords form angles $12^\circ$ and $18^\circ$ with common tangent, the total angle between $PC-PCC$ and $PCC-PT$ is $30^\circ$. 10. Use the law of cosines to find length $PC-PT$: $$PC-PT^2 = PC-PCC^2 + PCC-PT^2 + 2 \times PC-PCC \times PCC-PT \times \cos(180^\circ - 30^\circ)$$ Because chord $PC-PT$ is parallel to tangent, angle between chords is supplementary to 30$^\circ$, i.e., 150$^\circ$. Calculate $\cos 150^\circ = -\cos 30^\circ = -\frac{\sqrt{3}}{2} \approx -0.8660$. Substitute values: $$PC-PT^2 = 71.8^2 + 49.5^2 + 2 \times 71.8 \times 49.5 \times (-0.8660)$$ Calculate: $$71.8^2 = 5155.24, \quad 49.5^2 = 2450.25$$ $$2 \times 71.8 \times 49.5 \times (-0.8660) = -6155.32$$ Sum: $$PC-PT^2 = 5155.24 + 2450.25 - 6155.32 = 1450.17$$ So: $$PC-PT = \sqrt{1450.17} \approx 38.07\, m$$ Final answers: - Radius of first curve $R_1 \approx 150.8$ m - Chord length $PC-PT \approx 38.1$ m