Math / Trigonometry

Question9. Two forest fire stations, $P$ and $Q$ , are 20.0 km apart. A ranger at station $Q$ sees a fire 15.0 km away. If the angle between the line $P Q$ and the line from $P$ to the fire is $25^{\circ}$ , how far, to the nearest tenth of a kilometre, is station $P$ from the fire?

Studdy Solution

STEP 1

What is this asking? We need to find the distance from one fire station to a fire, knowing the distance between the stations, the distance from the other station to the fire, and the angle between them. Watch out! There might be two possible locations for the fire, so we might get two answers!
Also, remember your units: kilometers!

STEP 2

1. Draw a diagram
2. Use the cosine law

STEP 3

Let's visualize this!
Imagine fire station $P$ and fire station $Q$ are $\text{20 km}$ apart.
We can draw a line segment to represent this.

STEP 4

Now, there's a fire somewhere, and station $Q$ sees it $\text{15 km}$ away.
We can draw another line segment from $Q$ to represent the distance to the fire.
Let's call the fire $F$ .

STEP 5

We know the angle between the line $PQ$ and the line from $P$ to the fire $F$ is $25^\circ$ .
We can mark this angle on our diagram.
We're trying to find the length of $PF$ , the distance from station $P$ to the fire.

STEP 6

We have a triangle with sides $PQ = \text{20 km}$ , $QF = \text{15 km}$ , and angle $P = 25^\circ$ .
We want to find $PF$ , which we can call $d$ .
The cosine law is perfect for this!

STEP 7

The cosine law states: $QF^2 = PQ^2 + PF^2 - 2 \cdot PQ \cdot PF \cdot \cos(P)$ .
Let's plug in our values: $15^2 = 20^2 + d^2 - 2 \cdot 20 \cdot d \cdot \cos(25^\circ)$

STEP 8

This simplifies to: $225 = 400 + d^2 - 40 \cdot d \cdot \cos(25^\circ)$ $0 = d^2 - 40 \cdot \cos(25^\circ) \cdot d + 175$

STEP 9

Now, $\cos(25^\circ) \approx \textbf{0.9063}$ , so we have: $0 = d^2 - 40 \cdot 0.9063 \cdot d + 175$ $0 \approx d^2 - 36.252 \cdot d + 175$

STEP 10

This is a quadratic equation!
We can use the quadratic formula to solve for $d$ : $d = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a=1$ , $b = -36.252$ , and $c=175$ .

STEP 11

Plugging in the values, we get: $d = \frac{36.252 \pm \sqrt{(-36.252)^2 - 4 \cdot 1 \cdot 175}}{2 \cdot 1}$ $d = \frac{36.252 \pm \sqrt{1314.231504 - 700}}{2}$ $d = \frac{36.252 \pm \sqrt{614.231504}}{2}$ $d = \frac{36.252 \pm 24.78365}{2}$

STEP 12

This gives us two possible solutions: $d = \frac{36.252 + 24.78365}{2} \approx \frac{61.03565}{2} \approx 30.517825 \approx \textbf{30.5}$ $d = \frac{36.252 - 24.78365}{2} \approx \frac{11.46835}{2} \approx 5.734175 \approx \textbf{5.7}$

STEP 13

The two possible distances from station $P$ to the fire are approximately $\textbf{30.5 km}$ and $\textbf{5.7 km}$ .

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Question9. Two forest fire stations, $P$ and $Q$ , are 20.0 km apart. A ranger at station $Q$ sees a fire 15.0 km away. If the angle between the line $P Q$ and the line from $P$ to the fire is $25^{\circ}$ , how far, to the nearest tenth of a kilometre, is station $P$ from the fire?

Studdy Solution

STEP 1

STEP 2

1. Draw a diagram
2. Use the cosine law

STEP 3

Let's visualize this!
Imagine fire station $P$ and fire station $Q$ are $\text{20 km}$ apart.
We can draw a line segment to represent this.

STEP 4

Now, there's a fire somewhere, and station $Q$ sees it $\text{15 km}$ away.
We can draw another line segment from $Q$ to represent the distance to the fire.
Let's call the fire $F$ .

STEP 5

We know the angle between the line $PQ$ and the line from $P$ to the fire $F$ is $25^\circ$ .
We can mark this angle on our diagram.
We're trying to find the length of $PF$ , the distance from station $P$ to the fire.

STEP 6

We have a triangle with sides $PQ = \text{20 km}$ , $QF = \text{15 km}$ , and angle $P = 25^\circ$ .
We want to find $PF$ , which we can call $d$ .
The cosine law is perfect for this!

STEP 7

The cosine law states: $QF^2 = PQ^2 + PF^2 - 2 \cdot PQ \cdot PF \cdot \cos(P)$ .
Let's plug in our values: $15^2 = 20^2 + d^2 - 2 \cdot 20 \cdot d \cdot \cos(25^\circ)$

STEP 8

This simplifies to: $225 = 400 + d^2 - 40 \cdot d \cdot \cos(25^\circ)$ $0 = d^2 - 40 \cdot \cos(25^\circ) \cdot d + 175$

STEP 9

Now, $\cos(25^\circ) \approx \textbf{0.9063}$ , so we have: $0 = d^2 - 40 \cdot 0.9063 \cdot d + 175$ $0 \approx d^2 - 36.252 \cdot d + 175$

STEP 10

This is a quadratic equation!
We can use the quadratic formula to solve for $d$ : $d = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a=1$ , $b = -36.252$ , and $c=175$ .

STEP 11

Plugging in the values, we get: $d = \frac{36.252 \pm \sqrt{(-36.252)^2 - 4 \cdot 1 \cdot 175}}{2 \cdot 1}$ $d = \frac{36.252 \pm \sqrt{1314.231504 - 700}}{2}$ $d = \frac{36.252 \pm \sqrt{614.231504}}{2}$ $d = \frac{36.252 \pm 24.78365}{2}$

STEP 12

This gives us two possible solutions: $d = \frac{36.252 + 24.78365}{2} \approx \frac{61.03565}{2} \approx 30.517825 \approx \textbf{30.5}$ $d = \frac{36.252 - 24.78365}{2} \approx \frac{11.46835}{2} \approx 5.734175 \approx \textbf{5.7}$

STEP 13

The two possible distances from station $P$ to the fire are approximately $\textbf{30.5 km}$ and $\textbf{5.7 km}$ .

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Question9. Two forest fire stations, PPP and QQQ, are 20.0 km apart. A ranger at station QQQ sees a fire 15.0 km away. If the angle between the line PQP QPQ and the line from PPP to the fire is 25∘25^{\circ}25∘, how far, to the nearest tenth of a kilometre, is station PPP from the fire?

Studdy Solution

STEP 1

STEP 2

1. Draw a diagram2. Use the cosine law

STEP 3

Let's **visualize** this!Imagine fire station PPP and fire station QQQ are 20 km\text{20 km}20 km apart.We can draw a line segment to represent this.

STEP 4

Now, there's a fire somewhere, and station QQQ sees it 15 km\text{15 km}15 km away.We can draw another line segment from QQQ to represent the distance to the fire.Let's call the fire FFF.

STEP 5

We know the angle between the line PQPQPQ and the line from PPP to the fire FFF is 25∘25^\circ25∘.We can mark this angle on our diagram.We're trying to find the length of PFPFPF, the distance from station PPP to the fire.

STEP 6

We have a triangle with sides PQ=20 kmPQ = \text{20 km}PQ=20 km, QF=15 kmQF = \text{15 km}QF=15 km, and angle P=25∘P = 25^\circP=25∘.We want to find PFPFPF, which we can call ddd.The **cosine law** is perfect for this!

STEP 7

STEP 8

This simplifies to: 225=400+d2−40⋅d⋅cos⁡(25∘)225 = 400 + d^2 - 40 \cdot d \cdot \cos(25^\circ)225=400+d2−40⋅d⋅cos(25∘) 0=d2−40⋅cos⁡(25∘)⋅d+1750 = d^2 - 40 \cdot \cos(25^\circ) \cdot d + 1750=d2−40⋅cos(25∘)⋅d+175

STEP 9

Now, cos⁡(25∘)≈0.9063\cos(25^\circ) \approx \textbf{0.9063}cos(25∘)≈0.9063, so we have: 0=d2−40⋅0.9063⋅d+1750 = d^2 - 40 \cdot 0.9063 \cdot d + 1750=d2−40⋅0.9063⋅d+175 0≈d2−36.252⋅d+1750 \approx d^2 - 36.252 \cdot d + 1750≈d2−36.252⋅d+175

STEP 10

This is a **quadratic equation**!We can use the **quadratic formula** to solve for ddd: d=−b±b2−4ac2ad = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}d=2a−b±b2−4ac​​ where a=1a=1a=1, b=−36.252b = -36.252b=−36.252, and c=175c=175c=175.

STEP 11

STEP 12

STEP 13

The two possible distances from station PPP to the fire are approximately 30.5 km\textbf{30.5 km}30.5 km and 5.7 km\textbf{5.7 km}5.7 km.

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Question9. Two forest fire stations, $P$ and $Q$ , are 20.0 km apart. A ranger at station $Q$ sees a fire 15.0 km away. If the angle between the line $P Q$ and the line from $P$ to the fire is $25^{\circ}$ , how far, to the nearest tenth of a kilometre, is station $P$ from the fire?

1. Draw a diagram
2. Use the cosine law

Let's visualize this!
Imagine fire station $P$ and fire station $Q$ are $\text{20 km}$ apart.
We can draw a line segment to represent this.

Now, there's a fire somewhere, and station $Q$ sees it $\text{15 km}$ away.
We can draw another line segment from $Q$ to represent the distance to the fire.
Let's call the fire $F$ .

We know the angle between the line $PQ$ and the line from $P$ to the fire $F$ is $25^\circ$ .
We can mark this angle on our diagram.
We're trying to find the length of $PF$ , the distance from station $P$ to the fire.

We have a triangle with sides $PQ = \text{20 km}$ , $QF = \text{15 km}$ , and angle $P = 25^\circ$ .
We want to find $PF$ , which we can call $d$ .
The cosine law is perfect for this!

This simplifies to: $225 = 400 + d^2 - 40 \cdot d \cdot \cos(25^\circ)$ $0 = d^2 - 40 \cdot \cos(25^\circ) \cdot d + 175$

Now, $\cos(25^\circ) \approx \textbf{0.9063}$ , so we have: $0 = d^2 - 40 \cdot 0.9063 \cdot d + 175$ $0 \approx d^2 - 36.252 \cdot d + 175$

This is a quadratic equation!
We can use the quadratic formula to solve for $d$ : $d = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$ where $a=1$ , $b = -36.252$ , and $c=175$ .

The two possible distances from station $P$ to the fire are approximately $\textbf{30.5 km}$ and $\textbf{5.7 km}$ .