ropes ab and bc are two of the ropes used to support a tent. the two ropes are attached to a stake at b and…

Explanation:

Step1: Find vector of rope BC

First, find the coordinates of points B and C. Assuming B as the origin (0,0,0) for simplicity of vector - calculation. The coordinates of C can be determined from the given dimensions. If we consider the x - y - z axes as shown, the position vector of C relative to B, $\vec{r}{BC}$: From the figure, if we assume B is at the origin, and using the given lengths, the coordinates of C are (1.5,3,0). So, $\vec{r}{BC}=1.5\vec{i}+3\vec{j}+0\vec{k}$.

Step2: Find unit - vectors of stake directions

Let's assume the stake has a direction vector. We need to consider all possible directions of the stake. However, if we consider the geometry, we can assume the stake lies in the plane formed by the attachment points. Let's consider the two - dimensional case in the plane of the stake's cross - section at B. But a more general way is to consider the normal vectors to the surfaces related to the stake. Since the stake is likely to be vertical or have a well - defined orientation, we can assume a simple case where we consider the vectors along the edges of the stake's cross - section. Let's assume the stake has a rectangular cross - section at B. The vectors along the edges of the cross - section at B can be considered. Let $\vec{u}1$ and $\vec{u}2$ be two unit vectors along the edges of the stake's cross - section at B. From the detail of the stake at B, if we assume one edge is along the x - direction and the other along the y - direction of the stake's cross - section, and using the given lengths 0.08 m and 0.16 m, we can form unit vectors. But a more straightforward approach is to use the dot - product formula $\theta=\cos^{-1}\left(\frac{\vec{a}\cdot\vec{b}}{\vert\vec{a}\vert\vert\vec{b}\vert}\right)$. Let's assume the stake has a main axis vector $\vec{s}$. For simplicity, if we assume the stake is vertical (a common case), and we project the vector $\vec{r}{BC}$ onto the plane perpendicular to the stake's axis. The magnitude of $\vec{r}{BC}$ is $\vert\vec{r}_{BC}\vert=\sqrt{(1.5)^2 + 3^2+0^2}=\sqrt{2.25 + 9}=\sqrt{11.25}\approx3.354$ m.

Step3: Calculate the smallest angle

We know that $\cos\theta=\frac{\vert\vec{r}{BC}\cdot\vec{s}\vert}{\vert\vec{r}{BC}\vert\vert\vec{s}\vert}$. If we assume the stake is vertical, and we consider the projection of $\vec{r}{BC}$ onto the horizontal plane. The vector $\vec{r}{BC}$ has components in the x - y plane. The smallest angle $\theta$ between $\vec{r}{BC}$ and the stake (assuming vertical) is given by $\theta=\cos^{-1}\left(\frac{\sqrt{(1.5)^2+3^2}}{\sqrt{(1.5)^2 + 3^2+0^2}}\right)$. In fact, we can also use the fact that if we consider the right - triangle formed by the projection of $\vec{r}{BC}$ on the plane perpendicular to the stake's axis. Let's assume the stake is vertical. The vector $\vec{r}{BC}$ has a horizontal projection. The angle between $\vec{r}{BC}$ and the vertical (stake direction) is $\theta=\tan^{-1}\left(\frac{\sqrt{(1.5)^2 + 3^2}}{0}\right)$ (in a non - degenerate case, we consider the components). The smallest angle $\theta$ between $\vec{r}{BC}$ and the stake is found using the dot - product formula. Let the stake's direction vector be $\vec{k}$ (assuming vertical). $\vec{r}{BC}\cdot\vec{k} = 0$. The angle $\theta$ between $\vec{r}{BC}$ and the stake is $\theta=\cos^{-1}\left(\frac{\vert\vec{r}{BC}\cdot\vec{k}\vert}{\vert\vec{r}{BC}\vert\vert\vec{k}\vert}\right)$. The vector $\vec{r}{BC}=1.5\vec{i}+3\vec{j}+0\vec{k}$, $\vert\vec{r}{BC}\vert=\sqrt{1.5^2 + 3^2}=\sqrt{2.25+9}=\sqrt{11.25}$, $\vert\vec{k}\vert = 1$, and $\vec{r}{BC}\cdot\vec{k}=0$. The smallest angle is the angle between $\vec{r}{BC}$ and the vertical. We know that $\cos\theta=\frac{\vert\vec{r}{BC}\cdot\vec{k}\vert}{\vert\vec{r}{BC}\vert\vert\vec{k}\vert}=0$. So, $\theta = 90^{\circ}$. But if we consider the non - vertical cases of the stake, we use the general dot - product formula. Let's assume the stake has a direction vector $\vec{s}$ such that we can find the components of $\vec{r}{BC}$ along and perpendicular to $\vec{s}$. The dot - product of $\vec{r}{BC}$ and a vector along the stake's axis. If we assume the stake has a direction vector $\vec{s}$ with components $(s_x,s_y,s_z)$. $\vec{r}{BC}\cdot\vec{s}=1.5s_x + 3s_y+0s_z$. The magnitude of $\vec{r}{BC}$ is $\vert\vec{r}{BC}\vert=\sqrt{(1.5)^2+3^2}$. The smallest angle $\theta$ is given by $\theta=\cos^{-1}\left(\frac{\vert1.5s_x + 3s_y\vert}{\vert\vec{r}{BC}\vert\vert\vec{s}\vert}\right)$. If we assume the stake is vertical, the vector $\vec{r}{BC}$ lies in a horizontal plane (in the x - y plane projection), and the smallest angle between $\vec{r}{BC}$ and the stake (vertical) is $\theta=\cos^{-1}\left(\frac{0}{\vert\vec{r}{BC}\vert}\right)=90^{\circ}$. But if we consider the stake's cross - section vectors, let the vectors along the cross - section of the stake at B be $\vec{v}_1$ and $\vec{v}2$. The dot - product $\vec{r}{BC}\cdot\vec{v}1$ and $\vec{r}{BC}\cdot\vec{v}_2$ are calculated. The magnitude of $\vec{v}1$ and $\vec{v}2$ are 1 (unit vectors). The smallest angle $\theta$ between $\vec{r}{BC}$ and the stake is $\theta=\cos^{-1}\left(\frac{\min(\vert\vec{r}{BC}\cdot\vec{v}1\vert,\vert\vec{r}{BC}\cdot\vec{v}2\vert)}{\vert\vec{r}{BC}\vert}\right)$. Let's assume the stake has a rectangular cross - section at B with side lengths 0.08 m and 0.16 m. We can form two unit vectors $\vec{u}1=\frac{0.08\vec{i}+0\vec{j}+0\vec{k}}{\sqrt{(0.08)^2}}$ and $\vec{u}2=\frac{0\vec{i}+0.16\vec{j}+0\vec{k}}{\sqrt{(0.16)^2}}$. $\vec{r}{BC}\cdot\vec{u}1=\frac{1.5\times0.08}{\sqrt{(0.08)^2}} = 1.5$, $\vec{r}{BC}\cdot\vec{u}2=\frac{3\times0.16}{\sqrt{(0.16)^2}}=3$. $\vert\vec{r}{BC}\vert=\sqrt{(1.5)^2 + 3^2}=\sqrt{11.25}$. $\cos\theta_1=\frac{\vert\vec{r}{BC}\cdot\vec{u}1\vert}{\vert\vec{r}{BC}\vert\vert\vec{u}1\vert}$, $\cos\theta_2=\frac{\vert\vec{r}{BC}\cdot\vec{u}2\vert}{\vert\vec{r}{BC}\vert\vert\vec{u}2\vert}$. The smallest of these angles is $\theta=\cos^{-1}\left(\frac{\min(\vert\vec{r}{BC}\cdot\vec{u}1\vert,\vert\vec{r}{BC}\cdot\vec{u}2\vert)}{\vert\vec{r}{BC}\vert}\right)$. After calculation, $\theta=\cos^{-1}\left(\frac{1.5}{\sqrt{11.25}}\right)\approx63.43^{\circ}$

Answer:

$63.43$