$$ 1 \leftarrow $$ Find the angle between $$ u=\langle 4,7\rangle $$ and $$ v=\langle 8,8\rangle $$, to the nearest tenth of a degree, The angle between $$ u $$ and $$ v $$ is $$ \square^{\circ} $$, (Type an integer or a decimal. Round to the nearest tenth as needed.)

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Let $$ u = \langle 4,7 \rangle $$ and $$ v = \langle 8,8 \rangle $$. We use the dot product formula for the angle $$\theta$$ between two vectors:
$$
\cos\theta = \frac{u \cdot v}{\|u\| \, \|v\|}
$$

1. Compute the dot product:
$$
u \cdot v = 4 \times 8 + 7 \times 8 = 32 + 56 = 88
$$

2. Compute the magnitudes:
$$
\|u\| = \sqrt{4^2 + 7^2} = \sqrt{16 + 49} = \sqrt{65}
$$
$$
\|v\| = \sqrt{8^2 + 8^2} = \sqrt{64 + 64} = \sqrt{128} = 8\sqrt{2}
$$

3. Substitute into the formula:
$$
\cos\theta = \frac{88}{\sqrt{65} \times 8\sqrt{2}} = \frac{88}{8 \sqrt{130}} = \frac{11}{\sqrt{130}}
$$

4. Solve for $$\theta$$:
$$
\theta = \arccos\left(\frac{11}{\sqrt{130}}\right)
$$

5. Approximate the value:
$$\sqrt{130} \approx 11.4018$$, so
$$
\frac{11}{\sqrt{130}} \approx \frac{11}{11.4018} \approx 0.9643
$$
$$
\theta \approx \arccos(0.9643) \approx 15.9^\circ
$$

The angle between $$ u $$ and $$ v $$ is $$ 15.9^\circ $$.
The angle between $$ u $$ and $$ v $$ is $$ 15.9^\circ $$.

Find the angle between and , to the nearest tenth of a degree,
The angle between and is ,
(Type an integer or a decimal. Round to the nearest tenth as needed.)

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Find the angle between and , to the nearest tenth of a degree, The angle between and is , (Type an integer or a decimal. Round to the nearest tenth as needed.)