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Vector applications in trigonometry

Learning goals

  • Resolves vectors into components using trigonometric ratios.
  • Combines vector components to determine resultant magnitude and direction.
  • Uses inverse trigonometric functions to recover direction angles.
  • Interprets vector results in applied contexts.

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Vector applications in trigonometry 1
Find the boat's velocity direction, measured north of east, to the nearest tenth of a degree.
Figure for problem 551838

Hints

- Read the horizontal and vertical components from point \(P\). - Treat those components as the legs of a right triangle. - Which trigonometric ratio connects the vertical component, horizontal component, and direction angle?

Solution

1. The endpoint shows an eastward component of \(4\) and a northward component of \(3\). 2. Therefore, \(\tan\theta=\frac{3}{4}\). 3. Thus, \(\theta=\tan^{-1}\left(\frac{3}{4}\right)\approx36.9^\circ\).

Answer

\(36.9^\circ\) north of east
Vector applications in trigonometry 2
A force has magnitude \(12\,\text{N}\) and is directed \(60^\circ\) above the positive \(x\)-axis. Find its vertical component in exact form.

Hints

- Resolve the force into horizontal and vertical components. - Decide which trigonometric ratio connects the vertical component to the given magnitude and angle. - Keep the exact trigonometric value rather than converting to a decimal.

Solution

The vertical component is opposite the \(60^\circ\) angle, so \(F_y=12\sin 60^\circ=12\cdot\frac{\sqrt{3}}{2}=6\sqrt{3}\,\text{N}\).

Answer

\(6\sqrt{3}\,\text{N}\)
Vector applications in trigonometry 3
A \(20\,\text{N}\) force acts at an angle of \(30^\circ\) above the horizontal. Find the vertical component of the force.

Hints

- Identify which component is opposite the given angle. - Choose the trigonometric ratio that relates the opposite side to the hypotenuse. - The force magnitude is the hypotenuse of the component triangle.

Solution

1. The vertical component is opposite the \(30^\circ\) angle, so use \(F_y=F\sin\theta\). 2. \(F_y=20\sin 30^\circ=20\cdot\frac12=10\,\text{N}\).

Answer

\(10\,\text{N}\)
Vector applications in trigonometry 4
The diagram shows two force segments from point \(O\). Interpret segment \(OA\) as force \(\mathbf{F}_1\) directed from \(O\) to \(A\), and segment \(OB\) as force \(\mathbf{F}_2\) directed from \(O\) to \(B\). One grid unit represents \(1\,\text{N}\). Find the resultant \(\mathbf{R}=\mathbf{F}_1+\mathbf{F}_2\) in component form and find \(\|\mathbf{R}\|\).
Figure for problem 556115

Hints

- Read each force's horizontal and vertical components from its terminal point. - Add corresponding components. - Use the magnitude formula on the resultant.

Solution

1. From the diagram, \(\mathbf{F}_1=\langle3,0\rangle\) and \(\mathbf{F}_2=\langle0,4\rangle\). 2. Add components: \(\mathbf{R}=\langle3,4\rangle\). 3. Its magnitude is \(\|\mathbf{R}\|=\sqrt{3^2+4^2}=5\,\text{N}\).

Answer

\(\mathbf{R}=\langle3,4\rangle\,\text{N}\) and \(\|\mathbf{R}\|=5\,\text{N}\).
Vector applications in trigonometry 5
A force of \(8\,\text{N}\) points at \(120^\circ\), measured counterclockwise from the positive \(x\)-axis. Find its signed horizontal component in exact form.

Hints

- The horizontal component is measured along the \(x\)-axis. - What is the sign of an \(x\)-component pointing to the left? - Relate the adjacent component to the force's magnitude and direction angle.

Solution

1. The horizontal component is \(F_x=8\cos120^\circ\). 2. Since the force points left, \(\cos120^\circ=-\frac{1}{2}\). 3. Thus \(F_x=-4\,\text{N}\).

Answer

\(-4\,\text{N}\)
Vector applications in trigonometry 6
The graph shows a boat's velocity \(\mathbf{v}\). East and north are the positive axes, and each grid unit represents \(1\,\text{mph}\). Find its signed eastward and northward components, then write the velocity in component form.
Figure for problem 562228

Hints

- Read each signed coordinate from the endpoint of the directed segment. - Which horizontal direction is negative on this graph? - Use the horizontal coordinate first in component form.

Solution

1. The endpoint is four grid units left and two units up from the origin. 2. Left gives an eastward component of \(-4\,\text{mph}\), and up gives a northward component of \(2\,\text{mph}\). 3. Therefore, the velocity is \(\langle-4,2\rangle\,\text{mph}\).

Answer

Eastward component: \(-4\,\text{mph}\); northward component: \(2\,\text{mph}\); velocity: \(\langle-4,2\rangle\,\text{mph}\).
Vector applications in trigonometry 7
A force has magnitude \(F\) and points \(25^\circ\) west of north. Which expression gives its signed horizontal component: \(F\cos25^\circ\), \(-F\cos25^\circ\), \(F\sin25^\circ\), or \(-F\sin25^\circ\)? Explain your choice.

Hints

- Locate the angle's reference direction before choosing sine or cosine. - Which side of the right triangle is horizontal when the angle is measured from north? - Determine the sign from east-versus-west direction.

Solution

1. The reference angle is measured from north, so the horizontal component is opposite the angle and uses sine. 2. West is the negative \(x\)-direction, so the component must be negative. 3. The correct expression is \(-F\sin25^\circ\).

Answer

\(-F\sin25^\circ\). The angle is measured from north, making the horizontal component opposite; west makes its sign negative.
Vector applications in trigonometry 8
Two successive velocity vectors are shown. Each grid unit represents \(1\,\text{m/s}\). a) Find the resultant velocity vector from \(O\) to \(B\). b) Find its speed. c) Find its direction angle measured counterclockwise from the positive x-axis, to the nearest tenth of a degree.
Figure for problem 556085

Hints

- Read each directed segment by subtracting its initial point from its endpoint. - Head-to-tail vectors add to the displacement from the first initial point to the last endpoint. - Use the resultant components to find both magnitude and direction.

Solution

1. Reading the endpoints from the coordinate grid gives \(\overrightarrow{OA}=\langle 4,0\rangle\) and \(\overrightarrow{AB}=\langle 0,3\rangle\), in meters per second. 2. The resultant is \(\overrightarrow{OB}=\langle 4,3\rangle\,\text{m/s}\). 3. Its speed is \(\sqrt{4^2+3^2}=5\,\text{m/s}\). 4. Its direction angle satisfies \(\tan\theta=\frac{3}{4}\), so \(\theta\approx36.9^\circ\).

Answer

a) \(\langle 4,3\rangle\,\text{m/s}\) b) \(5\,\text{m/s}\) c) \(36.9^\circ\)
Vector applications in trigonometry 9
A rescue boat moves due north at \(12\,\text{mph}\) relative to the water while a current flows due east at \(5\,\text{mph}\). a) Find the boat's velocity relative to the shore in component form, using east as the positive x-direction and north as the positive y-direction. b) Find the boat's speed relative to the shore. c) Find the angle its path makes east of north, to the nearest tenth of a degree.

Hints

- Represent the boat velocity and current with perpendicular component vectors. - Add the component vectors before finding the speed. - Pay attention to whether the requested angle is measured from east or from north.

Solution

1. The boat's water-relative velocity is \(\langle 0,12\rangle\) and the current is \(\langle 5,0\rangle\), so the shore-relative velocity is \(\langle 5,12\rangle\,\text{mph}\). 2. The speed is \(\sqrt{5^2+12^2}=13\,\text{mph}\). 3. If \(\alpha\) is measured east of north, then \(\tan\alpha=\frac{5}{12}\). 4. Therefore, \(\alpha\approx22.6^\circ\).

Answer

a) \(\langle 5,12\rangle\,\text{mph}\) b) \(13\,\text{mph}\) c) \(22.6^\circ\) east of north
Vector applications in trigonometry 10
Two forces acting at \(O\) are shown. Each grid unit represents \(1\,\text{N}\). a) Find the resultant force vector. b) Find the magnitude of the resultant to the nearest tenth of a newton. c) Find the equilibrant vector, the single force that would balance the two shown forces.
Figure for problem 556091

Hints

- Read each force vector from its initial point and endpoint on the coordinate grid. - Add the two forces componentwise to obtain the resultant. - A balancing force must have the same magnitude as the resultant and the opposite direction.

Solution

1. Reading the endpoint coordinates from the diagram gives the two forces \(\langle 6,2\rangle\) and \(\langle -1,4\rangle\), in newtons. 2. Their resultant is \(\langle 6,2\rangle+\langle -1,4\rangle=\langle 5,6\rangle\,\text{N}\). 3. Its magnitude is \(\sqrt{5^2+6^2}=\sqrt{61}\approx7.8\,\text{N}\). 4. The equilibrant is the opposite of the resultant, so it is \(\langle -5,-6\rangle\,\text{N}\).

Answer

a) \(\langle 5,6\rangle\,\text{N}\) b) \(7.8\,\text{N}\) c) \(\langle -5,-6\rangle\,\text{N}\)
Vector applications in trigonometry 11
A drone flies at \(18\,\text{m/s}\) in a direction \(30^\circ\) east of north. Find its eastward and northward velocity components in exact form.

Hints

- Sketch a right triangle with north as the angle's reference side. - Decide which component is adjacent to that angle and which is opposite. - The given speed is the length of the velocity vector.

Solution

1. Because the angle is measured from north, the eastward component uses sine and the northward component uses cosine. 2. The eastward component is \(18\sin30^\circ=9\,\text{m/s}\). 3. The northward component is \(18\cos30^\circ=9\sqrt3\,\text{m/s}\).

Answer

Eastward: \(9\,\text{m/s}\); northward: \(9\sqrt3\,\text{m/s}\).
Vector applications in trigonometry 12
The graph shows two successive moves by a drone. Each grid unit represents \(1\,\text{m}\). a) Write the vectors \(\mathbf{u}\) and \(\mathbf{v}\) in component form, showing their signed components. b) Add the vectors component by component to find the net displacement. Show the addition, even though the final point is visible. c) Find the net displacement's direction angle measured counterclockwise from the positive \(x\)-axis, to the nearest tenth of a degree.
Figure for problem 562231

Hints

- Read each move as endpoint minus starting point. - Add the east-west movements and the north-south movements separately. - Before taking an inverse tangent, identify the quadrant of the total displacement.

Solution

1. The first move is \(\langle2,1\rangle\,\text{m}\), and the second is \(\langle-3,2\rangle\,\text{m}\). 2. Their sum is \(\langle-1,3\rangle\,\text{m}\). 3. The net vector points to quadrant II. Its direction angle is \(180^\circ-\tan^{-1}(3)\approx108.4^\circ\).

Answer

a) \(\mathbf{u}=\overrightarrow{OA}=\langle2,1\rangle\,\text{m}\); \(\mathbf{v}=\overrightarrow{AB}=\langle-3,2\rangle\,\text{m}\) b) \(\langle2,1\rangle+\langle-3,2\rangle=\langle2-3,1+2\rangle=\langle-1,3\rangle\,\text{m}\) c) \(108.4^\circ\)
Vector applications in trigonometry 13
A force has magnitude \(10\,\text{N}\). Its horizontal component is \(-5\,\text{N}\), and its vertical component is positive. Find its direction angle, measured counterclockwise from the positive \(x\)-axis.

Hints

- Use the horizontal component to find the cosine of the direction angle. - The signs of the two components tell you the quadrant. - Which angle has the required cosine in that quadrant?

Solution

1. The horizontal component is \(10\cos\theta=-5\), so \(\cos\theta=-\frac{1}{2}\). 2. A negative horizontal component and positive vertical component place the force in quadrant II. 3. The quadrant-II angle with this cosine is \(\theta=120^\circ\).

Answer

\(120^\circ\)
Vector applications in trigonometry 14
A swimmer wants to cross a river that is \(100\,\text{m}\) wide. The current is \(\mathbf{v}_w=\langle 0.6,0,0\rangle\,\text{m/s}\), where the positive \(x\)-axis runs parallel to the bank. The swimmer's speed relative to the water is \(1.0\,\text{m/s}\). a) The swimmer initially points straight across the river, so \(\mathbf{v}_s=\langle 0,1.0,0\rangle\,\text{m/s}\). Find the resulting velocity over the ground and its magnitude, rounded to the nearest hundredth of a meter per second. b) Find the crossing time and the downstream drift. c) Find the swimmer's velocity relative to the water if the swimmer wants to travel straight across with no downstream drift. d) Use a dot product to find the angle between the heading vector from part c) and the positive \(x\)-axis. Show the dot-product expression and round the angle to the nearest hundredth of a degree.

Hints

- Combine the swimming velocity and current as vectors. - Crossing time depends on the component perpendicular to the banks. - For no drift, the two components parallel to the bank must cancel. - Use the fixed swimming speed to determine the remaining component.

Solution

1. Add the swimmer and current velocities: \(\mathbf{v}_r=\mathbf{v}_s+\mathbf{v}_w=\langle 0.6,1.0,0\rangle\,\text{m/s}\). 2. Its magnitude is \(\|\mathbf{v}_r\|=\sqrt{0.6^2+1.0^2}=\sqrt{1.36}\approx1.17\,\text{m/s}\). 3. The across-river component is \(1.0\,\text{m/s}\), so the crossing time is \(\frac{100}{1.0}=100\,\text{s}\). The drift is \(0.6\cdot100=60\,\text{m}\). 4. To eliminate drift, the swimmer's \(x\)-component must be \(-0.6\,\text{m/s}\). With total speed \(1.0\,\text{m/s}\), the across-river component is \(\sqrt{1-0.6^2}=0.8\,\text{m/s}\). 5. Thus, \(\mathbf{v}_h=\langle -0.6,0.8,0\rangle\,\text{m/s}\). 6. Let \(\mathbf{i}=\langle 1,0,0\rangle\). Then \(\cos\alpha=\frac{\mathbf{v}_h\cdot\mathbf{i}}{\|\mathbf{v}_h\|\|\mathbf{i}\|}=-0.6\), so \(\alpha\approx126.87^\circ\).

Answer

a) \(\mathbf{v}_r=\langle 0.6,1.0,0\rangle\,\text{m/s}\), with magnitude \(\approx1.17\,\text{m/s}\) b) Crossing time: \(100\,\text{s}\); downstream drift: \(60\,\text{m}\) c) \(\mathbf{v}_h=\langle -0.6,0.8,0\rangle\,\text{m/s}\) d) \(\mathbf{v}_h\cdot\langle1,0,0\rangle=-0.6\), so \(\cos\alpha=\frac{-0.6}{1\cdot1}=-0.6\) and \(\alpha\approx126.87^\circ\) measured counterclockwise from the positive \(x\)-axis.
Vector applications in trigonometry 15
A survey aircraft flies \(120\,\text{mi}\) due east, then \(80\,\text{mi}\) at \(30^\circ\) north of east. Find the magnitude and direction of the aircraft's total displacement from its starting point. Give the magnitude to the nearest tenth of a mile and the direction to the nearest tenth of a degree north of east.

Hints

- Convert each leg of the trip into horizontal and vertical components. - Add corresponding components before finding the overall displacement. - Use the final component signs to identify the correct direction reference.

Solution

1. The first displacement is \(\langle 120,0\rangle\). 2. The second displacement is \(\langle 80\cos30^\circ,80\sin30^\circ\rangle=\langle 40\sqrt3,40\rangle\). 3. The total displacement is \(\langle 120+40\sqrt3,40\rangle\). 4. Its magnitude is \(\sqrt{(120+40\sqrt3)^2+40^2}\approx193.5\,\text{mi}\). 5. Its direction is \(\theta=\tan^{-1}\left(\frac{40}{120+40\sqrt3}\right)\approx11.9^\circ\) north of east.

Answer

\(193.5\,\text{mi}\) at \(11.9^\circ\) north of east
Vector applications in trigonometry 16
Two forces act on a ring. One force is \(50\,\text{N}\) due east. The second is \(40\,\text{N}\) at a direction angle of \(120^\circ\), measured counterclockwise from east. Find the magnitude and direction angle of the resultant force. Round both to the nearest tenth.

Hints

- Resolve the angled force into x- and y-components. - Add the two force vectors componentwise. - Use the signs of the resultant components before selecting the direction angle.

Solution

1. The first force is \(\langle 50,0\rangle\). 2. The second force is \(\langle 40\cos120^\circ,40\sin120^\circ\rangle=\langle -20,20\sqrt3\rangle\). 3. The resultant is \(\langle 30,20\sqrt3\rangle\). 4. Its magnitude is \(\sqrt{30^2+(20\sqrt3)^2}=10\sqrt{21}\approx45.8\,\text{N}\). 5. Both components are positive, so the resultant is in quadrant I. Its direction angle is \(\tan^{-1}\left(\frac{20\sqrt3}{30}\right)\approx49.1^\circ\).

Answer

\(45.8\,\text{N}\) at a direction angle of \(49.1^\circ\)
Vector applications in trigonometry 17
A small aircraft has an airspeed of \(100\,\text{mph}\). A wind blows due east at \(20\,\text{mph}\). The pilot wants the ground track to be due north. a) How many degrees west of north should the pilot point the aircraft? Round to the nearest tenth of a degree. b) What will the aircraft's ground speed be? Round to the nearest tenth of a mile per hour.

Hints

- The ground track is due north only if the east-west components cancel. - Use the wind speed to determine the required westward component of the air-velocity vector. - The airspeed is the magnitude of the aircraft's velocity relative to the air.

Solution

1. For the ground track to have no east-west component, the aircraft's air-velocity vector must have a westward component of \(20\,\text{mph}\) to cancel the wind. 2. If \(\theta\) is the heading angle west of north, then \(100\sin\theta=20\), so \(\theta=\sin^{-1}(0.2)\approx11.5^\circ\). 3. The northward air-velocity component is \(\sqrt{100^2-20^2}=40\sqrt6\approx98.0\,\text{mph}\). 4. The wind has no north-south component, so the ground speed is approximately \(98.0\,\text{mph}\).

Answer

a) \(11.5^\circ\) west of north b) \(98.0\,\text{mph}\)
Vector applications in trigonometry 18
A small airplane flies from airport \(A(0, 0, 0)\) to a destination \(B(0, 540, 0)\), measured in kilometers. A constant wind has velocity \(\mathbf{v}_w=\langle 80,-60,0\rangle\,\text{km/h}\). The airplane's airspeed is \(170\,\text{km/h}\). a) Find the ground-velocity vector \(\mathbf{v}_g\) required to travel directly along the positive \(y\)-axis. b) Find the corresponding air-velocity vector \(\mathbf{v}_h\). c) Use a dot product to find the angle between the airplane's heading and its ground track. Show the dot product and the cosine equation; round to the nearest hundredth of a degree. d) Find the flight time.

Hints

- A due-north ground velocity has zero \(x\)-component. - Relate ground velocity, air velocity, and wind velocity with vector addition. - Use the fixed airspeed to determine the unknown northward component. - Use distance divided by ground speed for the flight time.

Solution

1. Write the required ground velocity as \(\mathbf{v}_g=\langle 0,v_y,0\rangle\), where \(v_y>0\). 2. Since \(\mathbf{v}_g=\mathbf{v}_h+\mathbf{v}_w\), the air-velocity vector must be \(\mathbf{v}_h=\langle -80,v_y+60,0\rangle\). 3. Use the airspeed: \((-80)^2+(v_y+60)^2=170^2\). 4. This gives \((v_y+60)^2=22{,}500\). The northward solution is \(v_y+60=150\), so \(v_y=90\). 5. Therefore, \(\mathbf{v}_g=\langle 0,90,0\rangle\,\text{km/h}\) and \(\mathbf{v}_h=\langle -80,150,0\rangle\,\text{km/h}\). 6. For the angle \(\phi\), \(\cos\phi=\frac{\mathbf{v}_h\cdot\mathbf{v}_g}{\|\mathbf{v}_h\|\|\mathbf{v}_g\|}=\frac{15}{17}\). Thus, \(\phi\approx28.07^\circ\). 7. The flight time is \(\frac{540\,\text{km}}{90\,\text{km/h}}=6\,\text{h}\).

Answer

a) \(\mathbf{v}_g=\langle 0,90,0\rangle\,\text{km/h}\) b) \(\mathbf{v}_h=\langle -80,150,0\rangle\,\text{km/h}\) c) \(\mathbf{v}_h\cdot\mathbf{v}_g=13{,}500\), so \(\cos\phi=\frac{13{,}500}{170\cdot90}=\frac{15}{17}\) and \(\phi\approx28.07^\circ\). d) \(6\,\text{h}\)
Vector applications in trigonometry 19
A parallelepiped has a base formed by \(\mathbf{u}=\langle 3,0,4\rangle\) and \(\mathbf{v}=\langle 0,5,0\rangle\). A third edge vector \(\mathbf{w}\) has length \(6\) and makes a \(30^\circ\) angle with a normal direction to the base. a) Find the area of the base. b) Find the volume of the parallelepiped. c) Explain geometrically why calculating base area times perpendicular height gives the same volume no matter which face is chosen as the base.

Hints

- Check whether the two base vectors are perpendicular. - The height is the component of \(\mathbf{w}\) in the normal direction. - How are perpendicular height, base area, and volume related for this solid? - Imagine slicing the solid parallel to a base face. What can you say about the areas of those slices?

Solution

1. The base vectors are perpendicular because \(\mathbf{u}\cdot\mathbf{v}=0\). Their lengths are \(\|\mathbf{u}\|=5\) and \(\|\mathbf{v}\|=5\), so the base area is \(G=5\cdot5=25\) square units. 2. The component of \(\mathbf{w}\) perpendicular to the base is \(h=\|\mathbf{w}\|\cos30^\circ=6\left(\frac{\sqrt{3}}{2}\right)=3\sqrt{3}\). 3. Therefore, the volume is \(V=Gh=25\cdot3\sqrt{3}=75\sqrt{3}\) cubic units. 4. Imagine slicing the solid by planes parallel to a chosen base face. Every slice has the same area as that face: shifting the slice changes its position but not its shape. These equal-area slices fill the perpendicular height, giving base area times height. The same slicing argument works when a different pair of opposite faces is chosen as the base.

Answer

a) \(25\) square units b) \(75\sqrt{3}\) cubic units c) Slices parallel to any chosen base face all have that face's area. Stacking these slices across the perpendicular height gives the volume, so choosing a different face changes the area and height but not their product.
Vector applications in trigonometry 20
A \(500\,\text{lb}\) sign hangs motionless from two cables. The left cable makes a \(30^\circ\) angle above the horizontal, and the right cable makes a \(45^\circ\) angle above the horizontal. Find the tension in each cable. Round to the nearest tenth of a pound.

Hints

- Because the sign is motionless, the horizontal and vertical force components each balance separately. - Resolve each cable tension into horizontal and vertical components. - Use the horizontal balance equation to express one tension in terms of the other before using the vertical balance.

Solution

1. Let \(T_L\) and \(T_R\) be the left- and right-cable tensions. 2. Horizontal equilibrium gives \(T_L\cos30^\circ=T_R\cos45^\circ\). 3. Vertical equilibrium gives \(T_L\sin30^\circ+T_R\sin45^\circ=500\). 4. From the horizontal equation, \(T_R=T_L\frac{\cos30^\circ}{\cos45^\circ}=T_L\frac{\sqrt6}{2}\). 5. Substitute into the vertical equation: \(\frac12T_L+\frac{\sqrt2}{2}\left(T_L\frac{\sqrt6}{2}\right)=500\), so \(T_L=500(\sqrt3-1)\approx366.0\,\text{lb}\). 6. Then \(T_R\approx448.3\,\text{lb}\).

Answer

Left cable: \(366.0\,\text{lb}\) Right cable: \(448.3\,\text{lb}\)

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