Given sin(theta) = 0.80, find the angle theta in degrees and in radians.

Step-by-step solution

1. 1

   Apply the inverse sine to both sides. The operation that undoes sine is $\sin^{-1}$ (arcsin):

   $\theta=\sin^{-1}(0.80)$

   Its principal range is $-90^\circ\le\theta\le90^\circ$, so it returns the single acute angle here.

2. 2

   Check the input is legal. Sine never exceeds $1$ in absolute value, and $0.80$ is inside $[-1,1]$, so a real angle exists. (An input like $1.2$ would have no solution — worth checking before pressing a button.)

3. 3

   Predict the answer before computing. $\sin45^\circ\approx0.707$ and $\sin60^\circ\approx0.866$, and $0.80$ sits between them, so $\theta$ must be between $45^\circ$ and $60^\circ$. This estimate is what catches a calculator left in radian mode.

4. 4

   Evaluate in degrees.

   $\theta=\sin^{-1}(0.80)=53.1301^\circ\approx53.13^\circ$

   Squarely inside the predicted $45^\circ$–$60^\circ$ window ✓.

5. 5

   Convert to radians. Multiply by $\dfrac{\pi}{180}$:

   $53.1301\times\frac{\pi}{180}=0.9273\ \text{rad}$

   As a check, $0.9273$ is a little over $\tfrac{\pi}{4}=0.7854$ and under $\tfrac{\pi}{3}=1.0472$, matching the degree answer.

6. 6

   Recognise the special triangle behind it. $0.80=\tfrac{4}{5}$, so this is the familiar $3$-$4$-$5$ right triangle: the angle opposite the side of length $4$. Its cosine is $\tfrac35=0.6$ and its tangent $\tfrac43$, and indeed $\tan(53.13^\circ)=1.3333$ ✓. Verifying back, $\sin(0.9273)=0.80000$ ✓.