Exercises — Law of sines — proof and applications (ambiguous case)

You know two angles ($A,B$) and one side ($a$). Two angles + a side = the AAS/ASA family (here the side $a$ is not between the two angles, so it is AAS). AAS is always unique: two angles fix the shape completely (the third angle is $C=180^{\circ }-35^{\circ }-65^{\circ }=80^{\circ }$), and one side fixes the size. No ambiguity — no swinging side is involved because we are given no "loose" second side to swing.

You know two sides ($a,b$) and an angle ($A$) that is opposite one of them, not between the two sides. That is SSA — the ambiguous case ⚠️. Yes, there could be 0, 1, or 2 triangles. We would have to compare $a$ against $h=b\sin A$ to decide. (Recognition only — we do not resolve it yet; that is Level 3.)

No. Here the known angle sits between the two known sides — that is SAS. No side is paired with a known opposite angle, so the Law of Sines ratio has two unknowns and cannot start. Reach for the Law of Cosines first to find the third side, then the Law of Sines can finish.

What we do: pair each side with its opposite angle: $\frac{a}{\sin A}=\frac{b}{\sin B}$. Why: we know $a$ and its opposite angle $A$, so that ratio is fully known; the target $b$ has known opposite angle $B$. $b=\frac{a\sin B}{\sin A}=\frac{8\sin 60^{\circ }}{\sin 40^{\circ }}=\frac{8(0.86603)}{0.64279}\approx 10.78$ Sanity check: $60^{\circ }>40^{\circ }$ so $b>a$ (bigger angle faces bigger side): $10.78>8$ ✓.

What we do: first get $C=180^{\circ }-40^{\circ }-60^{\circ }=80^{\circ }$, then $\frac{c}{\sin C}=\frac{a}{\sin A}$. Why: we need an angle opposite $c$; the angle sum hands us $C$ for free. $c=\frac{a\sin C}{\sin A}=\frac{8\sin 80^{\circ }}{\sin 40^{\circ }}=\frac{8(0.98481)}{0.64279}\approx 12.26$ Sanity check: $C=80^{\circ }$ is the largest angle, so $c$ should be the longest side: $12.26>10.78>8$ ✓.

What we do: use the far-right piece of the law, $\frac{a}{\sin A}=2R$, so $R=\frac{a}{2\sin A}$. Why: the constant ratio is the circumdiameter (see Circumcircle and Circumradius R); one matched side–angle pair is enough. $R=\frac{10}{2\sin 50^{\circ }}=\frac{10}{2(0.76604)}=\frac{10}{1.53209}\approx 6.53$

Step 1 — critical height. $h=b\sin A=10\sin 30^{\circ }=10(0.5)=5$. Step 2 — classify. Order $h,a,b$: since $5<6<10$, i.e. $h<a<b$, the rule gives the two-triangle branch. Why: $\sin B=\frac{h}{a}=\frac{5}{6}<1$ has an acute and a supplementary solution, and $a<b$ lets both survive — the swinging tip crosses the base on both sides of the dotted foot in the figure. Step 3 — find $B$. $\sin B=\frac{b\sin A}{a}=\frac{10(0.5)}{6}=0.83333$. $B_1={\sin }^{-1}(0.83333)\approx 56.44^{\circ },B_2=180^{\circ }-56.44^{\circ }=123.56^{\circ }.$ Step 4 — keep both? Check angle sums: $30^{\circ }+56.44^{\circ }=86.44^{\circ }<180^{\circ }$ ✓ → $C_1=93.56^{\circ }$. $30^{\circ }+123.56^{\circ }=153.56^{\circ }<180^{\circ }$ ✓ → $C_2=26.44^{\circ }$. Both valid — two triangles.

Step 1 — height. $h=8\sin 50^{\circ }=8(0.76604)=6.13$. Step 2 — classify. $a=3<6.13=h$, i.e. the $a<h$ branch → 0 triangles. Why: the swinging orange side is shorter than the dotted vertical $h$, so its tip never reaches the base line. Cross-check with the sine: $\sin B=\frac{8\sin 50^{\circ }}{3}=\frac{6.128}{3}=2.043>1$, which is impossible for any angle — exactly what the $a<h$ branch predicts. Confirms no triangle.

Step 1 — height. $h=9\sin 45^{\circ }=9(0.70711)=6.36$. Step 2 — classify. Is $a$ above $h$ but below $b$? $6.36<7<9$ → $h<a<b$ → the two-triangle branch. Why: $\sin B=\frac{h}{a}=\frac{6.36}{7}<1$ has two solutions, and $a<b$ lets both pass the angle-sum test (the tip crosses twice). Step 3 — $B$. $\sin B=\frac{9\sin 45^{\circ }}{7}=\frac{6.364}{7}=0.90914$. $B_1={\sin }^{-1}(0.90914)\approx 65.36^{\circ }$, $B_2=180^{\circ }-65.36^{\circ }=114.64^{\circ }$. Step 4 — validity. $45^{\circ }+65.36^{\circ }=110.36^{\circ }<180^{\circ }$ ✓ ($C_1=69.64^{\circ }$); $45^{\circ }+114.64^{\circ }=159.64^{\circ }<180^{\circ }$ ✓ ($C_2=20.36^{\circ }$). Both valid.

Step 1 — classify. Here $a=12\ge b=9$, the $a\ge b$ branch → exactly one triangle. Why: since $a$ is the longer side and faces the given angle $A$, side $B$ cannot be obtuse (the biggest side faces the biggest angle), so the supplementary option is doomed before we even compute it. Step 2 — confirm. $\sin B=\frac{9\sin 60^{\circ }}{12}=\frac{7.794}{12}=0.64952$, so $B_1={\sin }^{-1}(0.64952)\approx 40.51^{\circ }$. The supplement $B_2=139.49^{\circ }$ gives $A+B_2=60^{\circ }+139.49^{\circ }=199.49^{\circ }>180^{\circ }$ — rejected. Only $B_1$ survives. ($C=79.49^{\circ }$.)

Step 1 — height. $h=b\sin A=8\sin 30^{\circ }=8(0.5)=4$. Step 2 — classify. Now $a=4=h$ exactly — the boundary $a=h$ branch → exactly one right-angled triangle. Why: $\sin B=\frac{h}{a}=\frac{4}{4}=1$, and the only angle with sine $1$ is $B=90^{\circ }$; the swinging orange side grazes the base tangentially at a single point (the dotted foot itself). Step 3 — finish. $B=90^{\circ }$, so $C=180^{\circ }-30^{\circ }-90^{\circ }=60^{\circ }$, and $c=\frac{a\sin C}{\sin A}=\frac{4\sin 60^{\circ }}{\sin 30^{\circ }}=\frac{4(0.86603)}{0.5}\approx 6.928$.

Plan: we need two sides and the included angle to use $\text{Area}=\frac{1}{2}ab\sin C$ (see Area of a Triangle). We have $a$; get $b$ by the Law of Sines; get $C$ from the angle sum. $C=180^{\circ }-70^{\circ }-60^{\circ }=50^{\circ }$. $b=\frac{a\sin B}{\sin A}=\frac{10\sin 60^{\circ }}{\sin 70^{\circ }}=\frac{10(0.86603)}{0.93969}\approx \mathrm{9.216.}$ Include the angle between $a$ and $b$: that angle is $C=50^{\circ }$. $\text{Area}=\frac{1}{2}ab\sin C=\frac{1}{2}(10)(9.216)\sin 50^{\circ }=\frac{1}{2}(10)(9.216)(0.76604)\approx \mathrm{35.30.}$

Why Law of Cosines first: SAS gives no matched side–angle pair, so the Law of Sines cannot start; Law of Cosines finds the missing side. $c^2=a^2+b^2-2ab\cos C=64+36-2(8)(6)\cos 40^{\circ }=100-96(0.76604)=100-73.54=26.46,$ $c=\sqrt{26.46}\approx \mathrm{5.144.}$ Now the Law of Sines can start (we have the pair $c,C$). We solve for angle $B$, whose side $b=6$ is not the longest, so $B$ is safely acute: $\sin B=\frac{b\sin C}{c}=\frac{6\sin 40^{\circ }}{5.144}=\frac{6(0.64279)}{5.144}=\frac{3.8567}{5.144}\approx 0.74975,$ $B\approx {\sin }^{-1}(0.74975)\approx 48.55^{\circ }.$ Cross-check via $2R$: $2R=\frac{c}{\sin C}=\frac{5.144}{0.64279}\approx 8.003$; and $\frac{b}{\sin B}=\frac{6}{0.74975}\approx 8.003$ ✓ — the ratios agree, confirming the arithmetic.

From L4.1: $a=10,b=9.216,C=50^{\circ }$, Area $\approx 35.30$. Find $c$ and $R$. $c=\frac{a\sin C}{\sin A}=\frac{10\sin 50^{\circ }}{\sin 70^{\circ }}=\frac{10(0.76604)}{0.93969}\approx \mathrm{8.152.}$ $2R=\frac{a}{\sin A}=\frac{10}{\sin 70^{\circ }}=\frac{10}{0.93969}\approx 10.642$, so $R\approx 5.321$. Check the identity: $\frac{abc}{4R}=\frac{10(9.216)(8.152)}{4(5.321)}=\frac{751.1}{21.28}\approx 35.30$ ✓ — matches the area from L4.1, tying Area of a Triangle to Circumcircle and Circumradius R.

Idea: replace each side by $2R$ times the sine of its opposite angle (that is the Law of Sines rearranged: $a=2R\sin A$, etc.). Then it becomes a pure trig identity. Right side: $b\cos C+c\cos B=2R\sin B\cos C+2R\sin C\cos B=2R(\sin B\cos C+\cos B\sin C).$ By the sine addition formula $\sin B\cos C+\cos B\sin C=\sin (B+C)$. But $B+C=180^{\circ }-A$, and $\sin (180^{\circ }-A)=\sin A$ (the supplementary-angle fact, see Sine of Supplementary Angles). So $b\cos C+c\cos B=2R\sin (B+C)=2R\sin (180^{\circ }-A)=2R\sin A=a.■$

Setup: the equation for $B$ is $\sin B=\frac{b\sin A}{a}$. If a value $s=\frac{b\sin A}{a}$ satisfies $0<s<1$, then two angles in $(0^{\circ },180^{\circ })$ have sine $s$: call them $B_1={\sin }^{-1}(s)$ (acute) and $B_2=180^{\circ }-B_1$. By construction $B_1+B_2=180^{\circ }$ — they are supplementary. ✓ Third angles: $C_1=180^{\circ }-A-B_1$ and $C_2=180^{\circ }-A-B_2=180^{\circ }-A-(180^{\circ }-B_1)=B_1-A$. Their difference: $C_1-C_2=(180^{\circ }-A-B_1)-(B_1-A)=180^{\circ }-2B_1.$ So $|C_1-C_2|=|180^{\circ }-2B_1|$. $■$ Interpretation: the two triangles share vertex angle $A$ and side ratios; they are the two positions of the swinging side in the figure, reflected about the foot of the altitude.

When $a=h=b\sin A$: substitute into $\sin B=\frac{b\sin A}{a}=\frac{a}{a}=1$. The only angle in $(0^{\circ },180^{\circ })$ with sine $1$ is $B=90^{\circ }$ — unique, so exactly one triangle. Why right-angled at $B$: $B=90^{\circ }$ means the side $a$ (opposite $B$) is the hypotenuse and coincides with the altitude direction — geometrically the swinging tip just grazes the base (tangent), touching at a single point. This is the knife-edge between "0 triangles" ($a<h$) and "2 triangles" ($h<a<b$). $■$