Answer:
[tex]\boxed{g\approx 2.0}[/tex]
Step-by-step explanation:
First, we can solve for the altitude using the trigonometric ratio cosine:
[tex]\cos(\theta)=\dfrac{\text{adjacent}}{\text{hypotenuse}}[/tex]
[tex]\cos(49\°) = \dfrac{a}{40}[/tex]
[tex]a=40\cos(49\°)[/tex]
[tex]a\approx12.02[/tex]
Next, we can solve for g using the trigonometric ratio tangent:
[tex]\tan(\theta)=\dfrac{\text{opposite}}{\text{adjacent}}[/tex]
[tex]\tan(63\°)=\dfrac{12.02}{g}[/tex]
[tex]\dfrac{1}{\tan(63\°)}=\dfrac{g}{12.02}[/tex]
[tex]g=12.02\tan(63\°)[/tex]
[tex]g\approx 2.04[/tex]
[tex]\boxed{g\approx 2.0}[/tex]
