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 \( \triangle \mathrm{ABC} \).

Alright, I need to solve this geometry problem involving triangle BDC and point A. Let's break down the information given and figure out how to approach each part of the question. First, let's summarize the given information: - AD is a vertical line. - Triangle BDC lies in a horizontal plane. - BC = 13 cm. - Angle DBC = 40 degrees. - Angle BDC = 110 degrees. - Angle of elevation of A from C is 35 degrees. - Angle of elevation of A from B is θ. - We need to calculate: 1. The length of AD. 2. The length of AB. 3. The magnitude of θ. 4. The magnitude of angle BAC. 5. The area of triangle ABC. Okay, let's start by visualizing the scenario. Since AD is vertical and triangle BDC is in a horizontal plane, point A must be somewhere above the plane, directly above point D. This means that AD is perpendicular to the plane of triangle BDC. Given that BC = 13 cm, and angles DBC and BDC are 40° and 110° respectively, we can deduce that angle B is 180° - 40° - 110° = 30°. So, angle B in triangle BDC is 30 degrees. Now, the angle of elevation of A from C is 35 degrees. This means that if we draw a line from point C to point A, the angle between this line and the horizontal plane is 35 degrees. Similarly, the angle of elevation of A from B is θ, which we need to find. Let's tackle the first part: calculating the length of AD. Since AD is vertical and forms a right angle with the horizontal plane, and the angle of elevation from C to A is 35 degrees, we can use trigonometry to find AD. The angle of elevation from C to A is the angle between the horizontal and the line AC. Therefore, in the right triangle formed by points A, C, and the projection of A on the plane, we have: tan(35°) = AD / CD But we don't know CD yet. Maybe we can find CD using the given angles in triangle BDC. In triangle BDC, we have: - BC = 13 cm - Angle B = 30° - Angle DBC = 40° - Angle BDC = 110° Using the Law of Sines in triangle BDC: sin(B) / BC = sin(DBC) / DC Plugging in the known values: sin(30°) / 13 = sin(40°) / DC Solving for DC: DC = (13 * sin(40°)) / sin(30°) Calculating the values: sin(30°) = 0.5 sin(40°) ≈ 0.6428 So, DC ≈ (13 * 0.6428) / 0.5 ≈ 16.72 cm Now, using the tangent of the angle of elevation from C to A: tan(35°) = AD / 16.72 Calculating tan(35°): tan(35°) ≈ 0.7002 So, AD ≈ 16.72 * 0.7002 ≈ 11.71 cm Therefore, the length of AD is approximately 11.71 cm. Next, let's find the length of AB. Since AD is vertical and AB is the hypotenuse of the right triangle ABD, we can use the Pythagorean theorem. First, we need to find BD. In triangle BDC, we have: - BC = 13 cm - DC ≈ 16.72 cm - Angle B = 30° Using the Law of Cosines to find BD: BD² = BC² + DC² - 2 * BC * DC * cos(B) Plugging in the values: BD² = 13² + 16.72² - 2 * 13 * 16.72 * cos(30°) BD² = 169 + 279.5584 - 2 * 13 * 16.72 * 0.8660 BD² = 448.5584 - 2 * 13 * 16.72 * 0.8660 BD² = 448.5584 - 374.24 BD² ≈ 74.3184 BD ≈ √74.3184 ≈ 8.62 cm Now, using the Pythagorean theorem in triangle ABD: AB² = AD² + BD² AB² = 11.71² + 8.62² AB² ≈ 137.0241 + 74.3184 AB² ≈ 211.3425 AB ≈ √211.3425 ≈ 14.56 cm So, the length of AB is approximately 14.56 cm. Now, let's find the magnitude of θ, the angle of elevation of A from B. We know that the angle of elevation from C to A is 35 degrees, and we've found AD ≈ 11.71 cm. To find θ, we can use the tangent function in triangle ABD. tan(θ) = AD / BD tan(θ) ≈ 11.71 / 8.62 ≈ 1.356 So, θ ≈ arctan(1.356) ≈ 53.5 degrees Therefore, the magnitude of θ is approximately 53.5 degrees. Next, let's find the magnitude of angle BAC. In triangle ABC, we have: - Angle at B = 30° - Angle at C = 35° (angle of elevation from C to A) - Angle at A = ? Since the sum of angles in a triangle is 180°: Angle at A = 180° - 30° - 35° = 115° So, the magnitude of angle BAC is 115 degrees. Finally, let's calculate the area of triangle ABC. We can use the formula for the area of a triangle: Area = (1/2) * base * height Here, we can consider BC as the base and AD as the height. Area = (1/2) * BC * AD Area = (1/2) * 13 * 11.71 ≈ 75.415 cm² Therefore, the area of triangle ABC is approximately 75.42 cm².