A volleyball player spikes the ball from a height of 2.44 meters. Assume that the path of the ball is a straight line. To the nearest degree, what is the maximum angle, θ, at which the ball can be hit and land within the court?
+33653 You have a right-angled triangle with opposite = 9.4m and adjacent = 2.44m. This means you use the tan function.
tan(θ) = 9.4/2.44 so θ = tan-1(9.4/2.44)
$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\frac{{\mathtt{9.4}}}{{\mathtt{2.44}}}}\right)} = {\mathtt{75.448\: \!623\: \!543\: \!84^{\circ}}}$$
so θ ≈ 75.4°
+33653 You have a right-angled triangle with opposite = 9.4m and adjacent = 2.44m. This means you use the tan function.
tan(θ) = 9.4/2.44 so θ = tan-1(9.4/2.44)
$$\underset{\,\,\,\,^{\textcolor[rgb]{0.66,0.66,0.66}{360^\circ}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\frac{{\mathtt{9.4}}}{{\mathtt{2.44}}}}\right)} = {\mathtt{75.448\: \!623\: \!543\: \!84^{\circ}}}$$
so θ ≈ 75.4°
