Practice questions

Chapter 5: Triangles and vectors

Try each problem first. Then check your work with the solution. Round to two decimal places.

1. The law of sines.

(a) Solve the triangle with ${A=35^\circ}$, ${B=75^\circ}$, and ${c=20}$.

(b) Solve the triangle with ${A=42^\circ}$, ${a=8}$, and ${b=10}$.

Solution:

(a)

The angles add up to ${180^\circ}$, so ${C=70^\circ}$. The known side ${c}$ is opposite ${C}$:

$a=\dfrac{20\sin 35^\circ}{\sin 70^\circ}\approx 12.21$

$b=\dfrac{20\sin 75^\circ}{\sin 70^\circ}\approx 20.56$

(b)

This is SSA, so check for two triangles:

$\sin B=\dfrac{10\sin 42^\circ}{8}\approx 0.8364$

So ${B\approx 56.76^\circ}$ or ${B\approx 123.24^\circ}$. Both leave room for ${C}$, so there are two triangles.

Triangle 1: ${B\approx 56.76^\circ}$ and ${C\approx 81.24^\circ}$. Then

$c=\dfrac{8\sin 81.24^\circ}{\sin 42^\circ}\approx 11.82$

Triangle 2: ${B\approx 123.24^\circ}$ and ${C\approx 14.76^\circ}$. Then

$c=\dfrac{8\sin 14.76^\circ}{\sin 42^\circ}\approx 3.05$

(c)

$\sin B=\dfrac{8\sin 60^\circ}{5}\approx 1.39$

A sine cannot be greater than ${1}$, so there is no such triangle.

2. The law of cosines.

(a) A triangle has ${b=6}$, ${c=9}$, and ${A=40^\circ}$. Find ${a}$.

(b) A triangle has sides ${5}$, ${6}$, and ${10}$. Find its largest angle.

(c) Two hikers leave camp along straight paths that make an angle of ${125^\circ}$. One walks ${3}$ mi, and the other walks ${5}$ mi. How far apart are they?

Solution:

(a)

$\begin{align*}a^2&=6^2+9^2-2(6)(9)\cos 40^\circ\\&\approx 117-82.73=34.27\end{align*}$

So ${a\approx\sqrt{34.27}\approx 5.85}$.

(b)

The largest angle ${C}$ is across from the longest side, ${10}$:

$\begin{align*}\cos C&=\dfrac{5^2+6^2-10^2}{2(5)(6)}\\&=\dfrac{-39}{60}=-0.65\end{align*}$

So ${C=\cos^{-1}(-0.65)\approx 130.54^\circ}$.

(c)

The paths are two sides with the angle ${125^\circ}$ between them:

$\begin{align*}&d^2\\&=3^2+5^2-2(3)(5)\cos 125^\circ\\&\approx 34+17.21=51.21\end{align*}$

So ${d\approx 7.16}$ mi.

3. Area of a triangle.

(a) Find the area of a triangle with sides ${10}$ and ${14}$ and an angle of ${35^\circ}$ between them.

(b) Find the area of a triangle with sides ${13}$, ${14}$, and ${15}$.

Solution:

(a)

$\begin{align*}\text{Area}&=\dfrac{1}{2}(10)(14)\sin 35^\circ\\&\approx 40.15\end{align*}$

(b)

Use Heron's formula. The semiperimeter is ${s=\dfrac{13+14+15}{2}=21}$, so ${s-a=8}$, ${s-b=7}$, ${s-c=6}$:

$\begin{align*}\text{Area}&=\sqrt{21\cdot 8\cdot 7\cdot 6}\\&=\sqrt{7056}=84\end{align*}$

4. Vectors.

(a) Find the vector from ${P(-2,3)}$ to ${Q(4,-5)}$, its magnitude, and the unit vector in its direction.

(b) Let ${\mathbf{u}=\langle 1,-2\rangle}$ and ${\mathbf{v}=\langle 3,4\rangle}$. Find ${3\mathbf{u}+2\mathbf{v}}$.

(d) A boat heads east at ${12}$ mi/h through water that flows north at ${5}$ mi/h. Find the boat's true speed and direction.

Solution:

(a)

$\begin{align*}\overrightarrow{PQ}&=\langle 4-(-2),\ -5-3\rangle\\&=\langle 6,-8\rangle\end{align*}$

${|\overrightarrow{PQ}|=\sqrt{36+64}=10}$

The unit vector is

$\dfrac{1}{10}\langle 6,-8\rangle=\left\langle\dfrac{3}{5},-\dfrac{4}{5}\right\rangle$

(b)

$\begin{align*}3\mathbf{u}+2\mathbf{v}&=\langle 3,-6\rangle+\langle 6,8\rangle\\&=\langle 9,2\rangle\end{align*}$

(c)

${210^\circ}$ is in quadrant III, with reference angle ${30^\circ}$:

$\begin{align*}&\langle 8\cos 210^\circ,\ 8\sin 210^\circ\rangle\\&=\left\langle 8\left(-\tfrac{\sqrt{3}}{2}\right),\ 8\left(-\tfrac{1}{2}\right)\right\rangle\\&=\langle -4\sqrt{3},\ -4\rangle\end{align*}$

(d)

With east along ${x}$ and north along ${y}$, the true velocity is

$\langle 12,0\rangle+\langle 0,5\rangle=\langle 12,5\rangle$

Speed: ${\sqrt{144+25}=13}$ mi/h.

Direction: ${\tan^{-1}\dfrac{5}{12}\approx 22.62^\circ}$ north of east.

5. The dot product.

(a) Find $\langle 5,-2\rangle\cdot\langle 3,4\rangle$.

(b) Find the angle between ${\langle 1,1\rangle}$ and ${\langle -1,\sqrt{3}\rangle}$.

(d) Find the projection of ${\mathbf{u}=\langle 2,6\rangle}$ onto ${\mathbf{v}=\langle 3,1\rangle}$.

(e) A force ${\mathbf{F}=\langle 4,3\rangle}$ (in newtons) moves an object from ${(1,1)}$ to ${(6,3)}$ (in meters). Find the work done.

Solution:

(a)

${5(3)+(-2)(4)=15-8=7}$

(b)

The dot product is ${-1+\sqrt{3}}$. The lengths are ${\sqrt{2}}$ and ${\sqrt{1+3}=2}$. So

$\cos\theta=\dfrac{\sqrt{3}-1}{2\sqrt{2}}\approx 0.2588$

and ${\theta=\cos^{-1}(0.2588)\approx 75^\circ}$. (In fact it is exactly ${75^\circ}$: the vectors have direction angles ${45^\circ}$ and ${120^\circ}$.)

(c)

${6(3)+(-9)(2)=18-18=0}$. Yes, they are perpendicular.

(d)

${\mathbf{u}\cdot\mathbf{v}=6+6=12}$, and ${|\mathbf{v}|^2=9+1=10}$:

$\text{proj}_{\mathbf{v}}\mathbf{u}=\dfrac{12}{10}\langle 3,1\rangle=\langle 3.6,\ 1.2\rangle$

(e)

The motion is

$\mathbf{D}=\langle 6-1,\ 3-1\rangle=\langle 5,2\rangle$

${W=\mathbf{F}\cdot\mathbf{D}=4(5)+3(2)=26}$ J

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