In the sketch alongside, AD is a vertical line. \( \triangle \mathrm{BDC} \) lies in a horizontal plane. \( \mathrm{BC}=13 \mathrm{~cm} \), \( \mathrm{DBC}=40^{\circ} . \mathrm{BDC}=110^{\circ} \). The angle of elevation of A from C is \( 35^{\circ} \) and the angle of elevation of A from B is \( \theta \). Calculate (1) the length of AD . (2) the length of AB . (3) the magnitude of \( \theta \). (4) the magnitude of BAC . (5) the area of \( \triangle \mathrm{ABC} \).

To solve the given geometric problem, we can break it down step by step. 1. **Calculate the length of AD:** We can use the Law of Sines in triangle BDC to find the lengths of the sides BC and DC. In triangle DBC, we know: - \( BC = 13 \, \text{cm} \) - \( \angle DBC = 40^\circ \) - \( \angle BDC = 110^\circ \) The remaining angle \( \angle BCD \) can be calculated as \( 180^\circ - (40^\circ + 110^\circ) = 30^\circ \). Using the Law of Sines: \[ \frac{BD}{\sin(30^\circ)} = \frac{BC}{\sin(110^\circ)} \] From this, we get \( BD = \frac{13 \cdot \sin(30^\circ)}{\sin(110^\circ)} \approx \frac{13 \cdot 0.5}{0.9397} \approx 6.93 \, \text{cm} \). Next, we find AD using the angle of elevation from C, \( \angle ACB = 35^\circ \): \[ \text{AD} = BC \cdot \tan(35^\circ) \approx 13 \cdot 0.7002 \approx 9.10 \, \text{cm}. \] 2. **Calculate the length of AB:** For triangle ABC, we can use the height AD and the length AC. We can use the tangent from point C to find AD. Using \( \tan(35^\circ) = \frac{AD}{AC} \), we get: \[ AC = \frac{AD}{\tan(35^\circ)} \approx \frac{9.10}{0.7002} \approx 12.99 \, \text{cm}. \] Using the cosine rule in triangle ABC where \( AB^2 = AC^2 + BC^2 - 2 \cdot AC \cdot BC \cdot \cos(30^\circ) \): \[ AB = \sqrt{(12.99)^2 + (13)^2 - 2 \cdot 12.99 \cdot 13 \cdot 0.866} \approx 12.99 \, \text{cm}. \] 3. **Magnitude of \( \theta \):** In triangle ABC, we already have values for sides AC and BC, including angle BAC: \[ \theta = \tan^{-1}\left(\frac{AD}{BC}\right) \approx \tan^{-1}\left(\frac{9.10}{13}\right) \approx 35.06^\circ. \] 4. **Magnitude of BAC:** Using the triangle angle sum property, \( \angle BAC = 180^\circ - (110^\circ + 30^\circ) = 40^\circ \). 5. **Area of triangle ABC:** The area can be calculated using the formula: \[ \text{Area} = \frac{1}{2} \cdot BC \cdot AD. \] Therefore, \[ \text{Area}_{ABC} = \frac{1}{2} \cdot 13 \cdot 9.10 \approx 59.15 \, \text{cm}^2. \] So, the results are: (1) Length of AD: \( \approx 9.10 \, \text{cm} \) (2) Length of AB: \( \approx 12.99 \, \text{cm} \) (3) Magnitude of \( \theta: \approx 35.06^\circ \) (4) Magnitude of BAC: \( 40^\circ \) (5) Area of triangle ABC: \( \approx 59.15 \, \text{cm}^2 \)