Solution of Triangles, Worked Examples (medium)

You have the complete side–angle pair $a$ and $A$, plus the side $b$, so start with the sine rule to find $\angle B$. Write it with the sines on top so the unknown angle is easy to isolate:

$$\dfrac{\sin B}{b}=\dfrac{\sin A}{a}\;\Rightarrow\; \sin B=\dfrac{b\sin A}{a}$$

Substitute $b = 6$, $A = 70^{\circ}$ and $a = 9$, with $\sin 70^{\circ}=0.9397$:

$$\sin B=\dfrac{6\sin 70^{\circ}}{9}=\dfrac{6\times0.9397}{9}=\dfrac{5.6382}{9}=0.6265$$

Take the inverse sine. Since $a = 9$ is longer than $b = 6$, angle $A$ is larger than $B$; with $A = 70^{\circ}$ that forces $B$ to be acute, so the acute value is the one we want:

$$B=\sin^{-1}(0.6265)=38.79^{\circ}$$

The three angles add to $180^{\circ}$, so $\angle C$ is what is left:

$$C=180^{\circ}-70^{\circ}-38.79^{\circ}=71.21^{\circ}$$

Now return to the sine rule for side $c$, pairing it with $C$:

$$c=\dfrac{a\sin C}{\sin A}=\dfrac{9\sin 71.21^{\circ}}{\sin 70^{\circ}}=\dfrac{9\times0.9467}{0.9397}=9.07$$

The area formula $\tfrac{1}{2}ab\sin C$ needs an angle, but you are only given sides, so recover an angle first with the cosine rule. Label the sides $a = 5$, $b = 6$ and $c = 7$; then the angle $C$ between $a$ and $b$ faces the side $c = 7$:

$$\cos C=\dfrac{a^{2}+b^{2}-c^{2}}{2ab}=\dfrac{5^{2}+6^{2}-7^{2}}{2(5)(6)}=\dfrac{25+36-49}{60}=\dfrac{12}{60}=0.2$$

This gives $C=\cos^{-1}(0.2)=78.46^{\circ}$. For the area you need $\sin C$; read it straight from the calculator, or from $\sin^{2}C+\cos^{2}C=1$:

$$\sin C=\sqrt{1-0.2^{2}}=\sqrt{0.96}=0.9798$$

Now apply the area formula with the two sides that enclose $C$, namely $a = 5$ and $b = 6$:

$$\text{Area}=\tfrac{1}{2}ab\sin C=\tfrac{1}{2}(5)(6)(0.9798)=15\times0.9798=14.70$$

Name the sides by the vertex opposite each: $BC = a$ faces $A$, $AC = b = 10$ faces $B$, and $AB = c = 6$ faces $C$. You have two sides and the included angle $A$, so the cosine rule gives $BC$:

$$a^{2}=b^{2}+c^{2}-2bc\cos A=10^{2}+6^{2}-2(10)(6)\cos 120^{\circ}$$

The angle is obtuse, so $\cos 120^{\circ}=-0.5$; subtracting a negative turns the middle term into an addition:

$$a^{2}=100+36-120(-0.5)=136+60=196\;\Rightarrow\; a=\sqrt{196}=14$$

So $BC = 14$ m. For the angle at $B$, use the sine rule, pairing $\angle B$ with the side it faces, $b = 10$:

$$\sin B=\dfrac{b\sin A}{a}=\dfrac{10\sin 120^{\circ}}{14}=\dfrac{10\times0.8660}{14}=0.6186$$

Because $a = 14$ is the longest side, $A = 120^{\circ}$ is the largest angle, so $B$ must be acute, take the acute value:

$$B=\sin^{-1}(0.6186)=38.21^{\circ}$$

Two angles are already known, so the third follows at once from the angle sum, and once all three angles are fixed, the sine rule connects the known side $AB$ to both unknown sides.

$$C = 180^{\circ} - 40^{\circ} - 75^{\circ} = 65^{\circ}$$

Side $AB = c$ faces $C$, so pair it with $C$ in the sine rule and solve for $a = BC$, the side opposite $A$:

$$a=\dfrac{c\sin A}{\sin C}=\dfrac{12\sin 40^{\circ}}{\sin 65^{\circ}}=\dfrac{12\times0.6428}{0.9063}=\dfrac{7.7136}{0.9063}=8.51$$

The same ratio, now paired with $B$, gives $b = AC$, the side opposite $B$:

$$b=\dfrac{c\sin B}{\sin C}=\dfrac{12\sin 75^{\circ}}{\sin 65^{\circ}}=\dfrac{12\times0.9659}{0.9063}=\dfrac{11.5908}{0.9063}=12.79$$

The area formula uses the angle enclosed by the two given sides, so rearrange it to isolate $\sin X$ first.

$$\text{Area}=\tfrac{1}{2}(XY)(XZ)\sin X\;\Rightarrow\;15=\tfrac{1}{2}(8)(5)\sin X=20\sin X$$

Solve for $\sin X$, then take the acute inverse sine, since the question asks for the acute angle:

$$\sin X=\dfrac{15}{20}=0.75\;\Rightarrow\;X=\sin^{-1}(0.75)=48.59^{\circ}$$

With $X$ known, the cosine rule, using the same two sides that enclose it, gives the third side $YZ$:

$$YZ^{2}=XY^{2}+XZ^{2}-2(XY)(XZ)\cos X=8^{2}+5^{2}-80(0.6614)=89-52.91=36.09\;\Rightarrow\;YZ=\sqrt{36.09}=6.01$$

The diagonal $AC$ splits the quadrilateral into two triangles that share that one side. Find $AC$ first, from triangle $ABC$, where two sides and the included angle are known:

$$AC^{2}=AB^{2}+BC^{2}-2(AB)(BC)\cos(\angle ABC)=6^{2}+8^{2}-2(6)(8)\cos100^{\circ}$$

$$AC^{2}=100-96(-0.1736)=100+16.67=116.67\;\Rightarrow\;AC=\sqrt{116.67}=10.80$$

This value of $AC$ now becomes a known side of triangle $ACD$, where all three sides are known. Apply the cosine rule again, this time to find $\angle ADC$, the angle facing $AC$:

$$\cos(\angle ADC)=\dfrac{CD^{2}+DA^{2}-AC^{2}}{2(CD)(DA)}=\dfrac{7^{2}+9^{2}-10.80^{2}}{2(7)(9)}=\dfrac{130-116.67}{126}=0.1058$$

$$\angle ADC=\cos^{-1}(0.1058)=83.93^{\circ}$$

Both bearings are measured from the same point $O$, so the angle between them is simply their difference:

$$\angle AOB = 100^{\circ}-30^{\circ}=70^{\circ}$$

Two sides and their included angle at $O$ are now known, so the cosine rule gives $AB$:

$$AB^{2}=OA^{2}+OB^{2}-2(OA)(OB)\cos(\angle AOB)=10^{2}+7^{2}-140\cos70^{\circ}=149-140(0.3420)=101.12$$

$$AB=\sqrt{101.12}=10.06$$

For the shortest distance from $O$ to line $AB$, write the area of triangle $OAB$ two ways, once from the two sides and included angle, once as $\tfrac{1}{2}\times AB\times h$, and equate them:

$$\text{Area}=\tfrac{1}{2}(OA)(OB)\sin(\angle AOB)=\tfrac{1}{2}(10)(7)\sin70^{\circ}=35(0.9397)=32.89$$

$$h=\dfrac{2\times\text{Area}}{AB}=\dfrac{2(32.89)}{10.06}=\dfrac{65.78}{10.06}=6.54$$

Let $AB = AC = x$. The cosine rule connects the base $BC$ to $x$ and the apex angle:

$$BC^{2}=x^{2}+x^{2}-2x^{2}\cos80^{\circ}=2x^{2}(1-0.1736)=1.6528x^{2}\;\Rightarrow\;BC=1.2856x$$

The perimeter gives a second equation linking $x$ and $BC$. Substitute the expression for $BC$ and solve for $x$:

$$2x+BC=32\;\Rightarrow\;2x+1.2856x=32\;\Rightarrow\;3.2856x=32\;\Rightarrow\;x=9.74$$

Substitute back to find $BC$:

$$BC=1.2856(9.74)=12.52$$