$\sin \alpha-1=0$ over the interval $0^{\circ} \leq \alpha \leq 180^{\circ}$, then $\alpha=$ ? (All angles are in degrees.)
Answer (1)
Given the equation sin α − 1 = 0 , we want to find α in the interval 0 ∘ ≤ α ≤ 18 0 ∘ .
Rewrite the equation as sin α = 1 .
Determine the angle α whose sine is 1, which is 9 0 ∘ .
The solution is 9 0 ∘ .
Explanation
Understanding the Problem We are given the equation sin α − 1 = 0 and asked to find the value of α in degrees within the interval 0 ∘ ≤ α ≤ 18 0 ∘ . The additional equations provided are irrelevant to the problem.
Isolating the Sine Function First, we isolate the sine function: sin α − 1 = 0 ⇒ sin α = 1
Finding the Angle Now, we need to find the angle α in the given interval whose sine is equal to 1. We know that sin 9 0 ∘ = 1 . Since the interval is 0 ∘ ≤ α ≤ 18 0 ∘ , the angle α = 9 0 ∘ is a valid solution.
Final Answer Therefore, the value of α that satisfies the given equation within the specified interval is 9 0 ∘ .
Examples
Imagine you're designing a ramp where the height is equal to the length of the base. The angle of elevation, α , can be found using trigonometric functions. If sin α = 1 , it means the ramp is at a 9 0 ∘ angle, forming a vertical wall rather than a ramp. This concept is crucial in engineering and construction to determine angles and slopes for various structures.
