on a flat field, jerry walks 80 m south from a to b. jerry then walks 150 m west from b to c. (a) state the bearing of b from a. (b) find the distance of ac. (c) find the bearing of c from a.

Question

Accepted Answer

### Part (a)
# Brief Explanations:
Bearing is measured clockwise from north. Since Jerry walks south from A to B, the bearing of B from A is $180^\circ$ (or S, but in standard bearing notation, it's $180^\circ$).
# Answer:
The bearing of B from A is $180^\circ$ (or "south" in non - standard bearing terms, but the standard 3 - figure bearing is $180^\circ$).

### Part (b)
# Explanation:
## Step 1: Identify the triangle type
The path from A to B to C forms a right - angled triangle, with $AB = 80\space m$ (vertical side, south direction) and $BC=150\space m$ (horizontal side, west direction). We can use the Pythagorean theorem to find $AC$. The Pythagorean theorem states that for a right - angled triangle with legs $a$ and $b$ and hypotenuse $c$, $c=\sqrt{a^{2}+b^{2}}$. Here, $a = 80$ and $b = 150$.
## Step 2: Apply the Pythagorean theorem
First, calculate $a^{2}+b^{2}$:
$a^{2}=80^{2}=6400$
$b^{2}=150^{2}=22500$
$a^{2}+b^{2}=6400 + 22500=28900$
Then, find the square root of $28900$:
$AC=\sqrt{28900}$
Since $170\times170 = 28900$, $\sqrt{28900}=170$
# Answer:
The distance of $AC$ is $170\space m$.

### Part (c)
# Explanation:
## Step 1: Find the angle $\theta$
We know that in the right - angled triangle $ABC$, $\tan\theta=\frac{BC}{AB}$, where $\theta$ is the angle between the south direction and the line $AC$ (measured from A).
We have $AB = 80\space m$ and $BC = 150\space m$. So $\tan\theta=\frac{150}{80}=\frac{15}{8}=1.875$
## Step 2: Calculate the angle $\theta$
Using the arctangent function, $\theta=\arctan(1.875)$. We know that $\arctan(1.875)\approx61.93^\circ$
## Step 3: Determine the bearing
The bearing of C from A is measured clockwise from north. The south direction is $180^\circ$ from north. We need to find the bearing by adding the angle $\theta$ to the south - west direction. The bearing is $180^\circ+ (90^\circ - 61.93^\circ)$? Wait, no. Let's re - think. The triangle is right - angled at B. So from point A, the line AC makes an angle $\theta$ with the south direction towards the west. So the bearing is $180^\circ+\theta$? No, wait. The standard bearing is measured from north. So the angle between north and the line AC: the south direction is $180^\circ$ from north. The angle between south and AC is $\theta=\arctan(\frac{150}{80})\approx61.93^\circ$ west of south. So to get the bearing from north, we go $180^\circ - 61.93^\circ$? No, no. Let's use the correct method.

The bearing is calculated as follows: The angle between the south line (from A) and AC is $\alpha=\arctan(\frac{BC}{AB})=\arctan(\frac{150}{80})\approx61.93^\circ$ (west of south). So the bearing from A to C is $180^\circ+ (90^\circ - 61.93^\circ)$? No, let's use the formula for bearing. The bearing is $180^\circ+\theta$, where $\theta$ is the angle between the south and AC towards west. Wait, actually, in terms of 3 - figure bearing:

We know that in triangle $ABC$, $\tan\theta=\frac{BC}{AB}=\frac{150}{80}=1.875$, so $\theta=\arctan(1.875)\approx61.93^\circ$

The bearing of C from A: we start at north, turn clockwise. The direction from A to C is south - west of some angle. The angle between the south direction and AC is $\theta\approx61.93^\circ$ west of south. So the bearing is $180^\circ+\theta$? No, the correct way is:

The bearing is $180^\circ+\ (90^\circ-\theta)$? No, let's calculate the angle from north. The south is $180^\circ$ from north. The angle between north and AC: the line AC is in the third quadrant (south - west). The angle between north and the negative y - axis (south) is $180^\circ$. The angle between the negative y - axis (south) and AC is $\theta=\arctan(\frac{150}{80})\approx61.93^\circ$ towards the negative x - axis (west). So the total bearing from north is $180^\circ+\ (90^\circ - 61.93^\circ)$? No, I think I made a mistake. Let's use the formula for bearing in a right - angled triangle.

The bearing of C from A is $180^\circ+\arctan(\frac{AB}{BC})$? No, $AB = 80$, $BC = 150$. $\tan\theta=\frac{AB}{BC}=\frac{80}{150}=\frac{8}{15}\approx0.533$, so $\theta=\arctan(\frac{8}{15})\approx28.07^\circ$. Wait, now I'm confused. Let's use the correct trigonometry.

In right - triangle $ABC$, right - angled at B. So angle at A: $\angle BAC$, $\tan(\angle BAC)=\frac{BC}{AB}=\frac{150}{80}=\frac{15}{8}$, so $\angle BAC=\arctan(\frac{15}{8})\approx61.93^\circ$

The bearing of C from A: we measure clockwise from north. The line AB is south (180 degrees from north). The angle between AB and AC is $\angle BAC\approx61.93^\circ$ towards the west. So the bearing is $180^\circ - 61.93^\circ$? No, that would be towards the east of south, which is wrong. Wait, no. The direction from A to C is south - west. So from north, we turn clockwise: north to east is 90, east to south is 90 (total 180), then south to west is some angle. Wait, the correct way is:

The bearing is $180^\circ+\ (90^\circ - 61.93^\circ)$? No, let's calculate the angle between north and AC. The vector from A to C is 80 south and 150 west. So in terms of standard position (from positive x - axis, but bearing is from positive y - axis (north) clockwise).

The x - component (west) is - 150 (if we take east as positive x), y - component (south) is - 80 (if we take north as positive y). The angle $\phi$ from the positive y - axis (north) clockwise is $180^\circ+\arctan(\frac{\vert x\vert}{\vert y\vert})$ when x is negative and y is negative.

$\vert x\vert = 150$, $\vert y\vert = 80$, $\arctan(\frac{150}{80})\approx61.93^\circ$

So the bearing is $180^\circ+61.93^\circ = 241.93^\circ\approx242^\circ$

# Answer:
The bearing of C from A is approximately $242^\circ$ (or more precisely $241.93^\circ$).

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QUESTION IMAGE

Question

Explanation:

Part (a)

Step 1: Identify the triangle type

Step 2: Apply the Pythagorean theorem

Step 1: Find the angle \(\theta\)

Step 2: Calculate the angle \(\theta\)

Step 3: Determine the bearing

Answer:

Part (b)