physics
A car is being pulled out of the mud by two forces that are applied by the two ropes shown in the drawing. The dashed line in the drawing bisects the 30.0° angle. The magnitude of the force applied by each rope is 2900 newtons (N). (a) How much force would a single rope …
Monday, January 27, 2014 at 8:38pm
physics
Three deer, A, B, and C, are grazing in a field. Deer B is located 63.4 m from deer A at an angle of 50.8 ° north of west. Deer C is located 76.7 ° north of east relative to deer A. The distance between deer B and C is 96.7 m. What is the distance between deer A and C?
Monday, January 27, 2014 at 8:36pm
The correct answer and explanation is:
Let’s solve the second problem:
Physics Problem (Deer A, B, and C):
Three deer, A, B, and C, are grazing in a field.
- Deer B is located 63.4 m from A, at an angle of 50.8° north of west.
- Deer C is located at an angle of 76.7° north of east relative to A.
- The distance between deer B and C is 96.7 m.
Question: What is the distance between deer A and C?
Step-by-step Explanation (300+ words):
We are given a triangle formed by deer A, B, and C. We can solve this problem using the Law of Cosines, but first, we need to understand the angles and construct a proper triangle.
Step 1: Understand the geometry
Let’s place Deer A at the origin (0, 0).
- Deer B is at 63.4 m at 50.8° north of west ⇒ this means 180° − 50.8° = 129.2° from the east (standard counterclockwise from east).
- Deer C is at some distance (unknown) at 76.7° north of east, which is simply 76.7° from the east.
Thus, we can calculate the angle ∠BAC between vectors A→B and A→C using: Angle at A=76.7°+50.8°=127.5°\text{Angle at A} = 76.7° + 50.8° = 127.5°
Step 2: Apply the Law of Cosines
Let:
- AB=63.4 mAB = 63.4 \, \text{m}
- BC=96.7 mBC = 96.7 \, \text{m}
- ∠BAC = 127.5°
- We want to find AC=xAC = x
From the Law of Cosines: BC2=AB2+AC2−2(AB)(AC)cos(∠BAC)BC^2 = AB^2 + AC^2 – 2(AB)(AC)\cos(\angle BAC)
Substitute: 96.72=63.42+x2−2(63.4)(x)cos(127.5°)96.7^2 = 63.4^2 + x^2 – 2(63.4)(x)\cos(127.5°)
Evaluate the cosine: cos(127.5°)≈−0.6018\cos(127.5°) ≈ -0.6018
Now plug in: 9355.69=4012.36+x2−2(63.4)(x)(−0.6018)9355.69 = 4012.36 + x^2 – 2(63.4)(x)(-0.6018) 9355.69=4012.36+x2+76.278×9355.69 = 4012.36 + x^2 + 76.278x
Rewriting: x2+76.278x+4012.36−9355.69=0x^2 + 76.278x + 4012.36 – 9355.69 = 0 x2+76.278x−5343.33=0x^2 + 76.278x – 5343.33 = 0
Solve using the quadratic formula: x=−76.278±76.2782+4(5343.33)2x = \frac{-76.278 \pm \sqrt{76.278^2 + 4(5343.33)}}{2} x=−76.278±5828.33+21373.322x = \frac{-76.278 \pm \sqrt{5828.33 + 21373.32}}{2} x=−76.278±27201.652x = \frac{-76.278 \pm \sqrt{27201.65}}{2} x=−76.278±164.932x = \frac{-76.278 \pm 164.93}{2}
Taking the positive root: x=88.652=44.33 mx = \frac{88.65}{2} = 44.33 \, \text{m}
✅ Final Answer:
The distance between deer A and C is approximately 44.3 me