You are climbing a mountain by the steepest route at a slope of 30° when you come upon a trail branching off at a 10° angle from yours

You are climbing a mountain by the steepest route at a slope of 30° when you come upon a trail branching off at a 10° angle from yours. What is the angle of ascent of the branch trail? angle = (in degrees)

The correct answer and explanation is:

Correct Answer: ≈ 5.2°

✅ Explanation (300 words)

To solve this problem, we need to find the angle of ascent of the branch trail, which makes a 10° angle with your current steepest path that has an inclination (slope) of 30°.

This is a 3D geometry problem involving vector components and trigonometry. Here’s how to think about it:

🧭 Understanding the Situation:

1. You’re climbing a mountain at a 30° slope.
2. A new trail branches off at an angle of 10° to your current path.
3. The branch trail does not go directly uphill anymore but at a 10° angle from the steepest route.
4. We’re asked: How steep (how much vertical gain) is the new trail? That is, what is its angle of ascent (with respect to the horizontal)?

📐 Step-by-step Solution:

Let’s use vector components:

• Let your original trail (slope = 30°) be along a direction that has:
  • A horizontal component: cos(30°)
  • A vertical component: sin(30°)

Now, the branch trail is at a 10° angle from the original path, but we assume it veers horizontally (not vertically). That means it is deviating from the steepest route horizontally, like walking diagonally across a hill.

So we calculate the vertical component of the branch trail’s direction: Vertical component=sin⁡(30°)⋅cos⁡(10°)\text{Vertical component} = \sin(30°) \cdot \cos(10°) =0.5⋅cos⁡(10°)≈0.5⋅0.9848≈0.4924= 0.5 \cdot \cos(10°) ≈ 0.5 \cdot 0.9848 ≈ 0.4924

The horizontal component: Horizontal component=1−(0.4924)2≈1−0.2425≈0.7575≈0.8702\text{Horizontal component} = \sqrt{1 – (0.4924)^2} ≈ \sqrt{1 – 0.2425} ≈ \sqrt{0.7575} ≈ 0.8702

Now, find the angle of ascent (θ): θ=tan⁡−1(verticalhorizontal)=tan⁡−1(0.49240.8702)≈tan⁡−1(0.5657)≈29.5∘\theta = \tan^{-1}\left(\frac{\text{vertical}}{\text{horizontal}}\right) = \tan^{-1}\left(\frac{0.4924}{0.8702}\right) ≈ \tan^{-1}(0.5657) ≈ \boxed{29.5^\circ}

Wait! That’s not quite right—let’s redo assuming the 10° is between the horizontal projections of the trails:

Instead, the proper formula is: sin⁡(θ)=sin⁡(30°)⋅cos⁡(10°)\sin(\theta) = \sin(30°) \cdot \cos(10°) θ=sin⁡−1(0.5⋅cos⁡(10°))=sin⁡−1(0.5⋅0.9848)≈sin⁡−1(0.4924)\theta = \sin^{-1}(0.5 \cdot \cos(10°)) = \sin^{-1}(0.5 \cdot 0.9848) ≈ \sin^{-1}(0.4924) θ≈29.5∘\theta ≈ \boxed{29.5^\circ}

So the correct answer is ≈ 29.5°, not 5.2°. That earlier assumption was wrong.

✅ Final Answer: 29.5° (degrees)

The branch trail still ascends steeply, just slightly less than the original trail.