Exercises — Area of triangle = ½ab·sin C

WHAT: Two sides and their included angle — direct substitution. WHY: $C=90^{\circ }$ means the two sides meet at a right angle, so they act as base and height directly. $\text{Area}=\frac{1}{2}(10)(4)\sin 90^{\circ }=\frac{1}{2}(40)(1)=20.$ Sanity check: with a right angle the sides are the legs, so area $=\frac{1}{2}\times \text{leg}\times \text{leg}=\frac{1}{2}(10)(4)=20$. ✓ Answer: $20$ square units.

WHAT: Substitute, keeping $\sin 60^{\circ }=\frac{\sqrt{3}}{2}$ exact. WHY exact: "Leave it exact" means don't round — carry the surd. $\text{Area}=\frac{1}{2}(6)(6)\left(\frac{\sqrt{3}}{2}\right)=\frac{1}{2}(36)\frac{\sqrt{3}}{2}=9\sqrt{3}\approx \mathrm{15.588.}$ Answer: $9\sqrt{3}\approx 15.59$ square units. (This is an equilateral triangle — all sides $6$, all angles $60^{\circ }$.)

WHAT: Treat the formula as an equation and solve for $a$. WHY: We know Area, $b$, and $\sin C$; only $a$ is unknown — one equation, one unknown. $24=\frac{1}{2}a(10)(0.8)=4a⟹a=\frac{24}{4}=6.$ Answer: $a=6$.

WHAT: Solve $\sin C$ first, then undo the sine. WHY two answers: $\sin$ takes the same value at an acute and its supplementary obtuse angle (this is the ambiguous case — see Unit Circle and Sine). $10=\frac{1}{2}(5)(8)\sin C=20\sin C⟹\sin C=\mathrm{0.5.}$ $C=30^{\circ }\text{or}C=180^{\circ }-30^{\circ }=150^{\circ }.$ Answer: $C=30^{\circ }$ or $C=150^{\circ }$. Both build a real triangle; extra context (e.g. "$C$ is acute") would pick one.

WHAT: The formula needs two sides and their included angle. Here we only have one side. So first grab a second side using the Sine Rule. WHY this route: Side $c$ sits between angles $A$ and $B$. If we also knew, say, side $b$, then $b$ and $c$ would enclose angle $A$, and we could use $\frac{1}{2}bc\sin A$. Step 1 — find the third angle (angles sum to $180^{\circ }$): $C=180^{\circ }-40^{\circ }-75^{\circ }=65^{\circ }.$ Step 2 — Sine Rule to get side $b$ (opposite $B$), knowing $c$ (opposite $C$): $\frac{b}{\sin B}=\frac{c}{\sin C}⟹b=\frac{9\sin 75^{\circ }}{\sin 65^{\circ }}=\frac{9(0.96593)}{0.90631}\approx \mathrm{9.5920.}$ Step 3 — area with the included angle $A$ (sides $b$ and $c$ enclose $A$): $\text{Area}=\frac{1}{2}bc\sin A=\frac{1}{2}(9.5920)(9)\sin 40^{\circ }=\frac{1}{2}(86.328)(0.64279)\approx \mathrm{27.75.}$ Answer: area $\approx 27.75$ square units.

WHAT: With $a,b$ fixed, area $=\frac{1}{2}(7)(7)\sin C=24.5\sin C$. So area is largest exactly when $\sin C$ is largest. WHY: $\sin C$ is the only moving part. On $0^{\circ }<C<180^{\circ }$, $\sin C$ peaks at $C=90^{\circ }$ where $\sin C=1$ (look at the top of the Unit Circle and Sine arc). $\text{Max area}=24.5(1)=24.5\text{at }C=90^{\circ }.$ What it looks like: the two equal poles open up until they're perpendicular — a right-angled isosceles triangle — then closing further shrinks the enclosed region. Answer: maximum area $24.5$ at $C=90^{\circ }$.

WHAT: Two ways to compute the same area let us back out the height on a different base. Step 1 — area from the included angle: $\text{Area}=\frac{1}{2}(8)(5)\sin 150^{\circ }=\frac{1}{2}(40)(0.5)=10.$ Step 2 — find side $c$ with the Cosine Rule ($\cos 150^{\circ }=-\frac{\sqrt{3}}{2}\approx -0.86603$): $c^2=a^2+b^2-2ab\cos C=64+25-2(8)(5)(-0.86603)=89+69.282=\mathrm{158.282.}$ $c=\sqrt{158.282}\approx \mathrm{12.581.}$ Step 3 — height on base $c$. Since area $=\frac{1}{2}\times c\times h_c$, $h_c=\frac{2\text{Area}}{c}=\frac{2(10)}{12.581}\approx \mathrm{1.590.}$ WHY this works: area is one fixed number no matter which side you call the base — so once you know the area and a chosen base, its perpendicular height is forced. Answer: area $=10$, and $h_c\approx 1.59$.

Part (a) — Heron. Semi-perimeter $s=\frac{6+8+10}{2}=12$. $\text{Area}=\sqrt{s(s-a)(s-b)(s-c)}=\sqrt{12\cdot 6\cdot 4\cdot 2}=\sqrt{576}=24.$ Part (b) — recover angle $C$. Set the area formula equal to $24$: $24=\frac{1}{2}(6)(8)\sin C=24\sin C⟹\sin C=1⟹C=90^{\circ }.$ Do they agree? Yes — and beautifully: $6^2+8^2=36+64=100=10^2$, so by Pythagoras this is a right triangle with the right angle exactly between $a$ and $b$. Both methods say area $=24$, angle $=90^{\circ }$. ✓ Answer: area $=24$, $C=90^{\circ }$.

Setup. From $P$, the two edge vectors are $\vec{u}=\overrightarrow{PQ}=(6,0),\vec{v}=\overrightarrow{PR}=(2,5).$ Part (a) — cross product. In 2D, the cross product's magnitude is $|u_x{v}_y-u_y{v}_x|$: $|\vec{u}\times \vec{v}|=|(6)(5)-(0)(2)|=|30|=30.$ Area $=\frac{1}{2}|\vec{u}\times \vec{v}|=\frac{1}{2}(30)=15.$ WHY $\frac{1}{2}$: the cross product gives the area of the whole parallelogram spanned by the two vectors; a triangle is half of it. Part (b) — area formula. The side lengths at $P$ are $a=|\vec{u}|=6,b=|\vec{v}|=\sqrt{2^2+5^2}=\sqrt{29}\approx \mathrm{5.3852.}$ The included angle $C$ (at $P$) has $\sin C=\frac{|\vec{u}\times \vec{v}|}{|\vec{u}||\vec{v}|}=\frac{30}{6\sqrt{29}}=\frac{5}{\sqrt{29}}$. $\text{Area}=\frac{1}{2}(6)(\sqrt{29})\cdot \frac{5}{\sqrt{29}}=\frac{1}{2}(6)(5)=15.$ The $\sqrt{29}$ cancels — the two methods are literally the same statement. This is why the parent note calls the cross product the vector twin of $\frac{1}{2}ab\sin C$. Answer: area $=15$ (both ways).

Step 1 — solve for $\sin C$: $27=\frac{1}{2}(9)(12)\sin C=54\sin C⟹\sin C=\mathrm{0.5.}$ Step 2 — pick the obtuse root. $\sin C=0.5$ gives $C=30^{\circ }$ or $150^{\circ }$. We're told $C$ is obtuse, so $C=150^{\circ }.$ Step 3 — third side via Cosine Rule ($\cos 150^{\circ }=-\frac{\sqrt{3}}{2}\approx -0.86603$): $c^2=9^2+12^2-2(9)(12)\cos 150^{\circ }=81+144-216(-0.86603)=225+187.06=\mathrm{412.06.}$ $c=\sqrt{412.06}\approx \mathrm{20.30.}$ Answer: $C=150^{\circ }$, $c\approx 20.30$.

Step 1 — one sine, two angles. $12=\frac{1}{2}(5)(6)\sin C=15\sin C⟹\sin C=\mathrm{0.8.}$ $C_1={\sin }^{-1}(0.8)\approx 53.13^{\circ }(\text{acute}),C_2=180^{\circ }-53.13^{\circ }=126.87^{\circ }(\text{obtuse}).$ Step 2 — third side for each via the Cosine Rule. Note $\cos 53.13^{\circ }=0.6$ and $\cos 126.87^{\circ }=-0.6$. For $C_1$: $c_1^2=25+36-2(5)(6)(0.6)=61-36=25⟹c_1=5.$ For $C_2$: $c_2^2=25+36-2(5)(6)(-0.6)=61+36=97⟹c_2=\sqrt{97}\approx \mathrm{9.849.}$ The insight: same two sides, same area, yet the shapes differ — one is a compact triangle ($c_1=5$), the other stretched wide ($c_2\approx 9.85$). Equal area does not mean equal triangle; the ambiguous case creates two genuinely different figures. Answer: $C_1\approx 53.13^{\circ }$, $c_1=5$; $C_2\approx 126.87^{\circ }$, $c_2\approx 9.85$.

Part (a). Both radii equal $r=10$, and $\theta$ is the included angle between them, so $\text{Area}(△OAB)=\frac{1}{2}(10)(10)\sin \theta =50\sin \theta .$ Part (b). Set this to $40$: $50\sin \theta =40⟹\sin \theta =\mathrm{0.8.}$ In radians on $0<\theta <\pi$, this gives two solutions (acute and obtuse), symmetric about $\frac{\pi }{2}$: $\theta ={\sin }^{-1}(0.8)\approx 0.9273\text{rad}\text{or}\theta =\pi -0.9273\approx 2.2143\text{rad}.$ WHY radians here: the mention of a circle/sector signals radian territory; make sure the calculator is in radian mode (a parent-note mistake!). Answer: $\text{Area}=50\sin \theta$; and $\theta \approx 0.927$ rad or $\theta \approx 2.214$ rad.

L1: $20$; $9\sqrt{3}$ L2: $a=6$; $C=30^{\circ }$ or $150^{\circ }$; area $\approx 27.75$ L3: max area $24.5$ at $C=90^{\circ }$; area $=10$, $h_c\approx 1.59$ L4: area $=24$, $C=90^{\circ }$; area $=15$ L5: $C=150^{\circ }$, $c\approx 20.30$; $C_1\approx 53.13^{\circ },c_1=5$ and $C_2\approx 126.87^{\circ },c_2\approx 9.85$; $\theta \approx 0.927$ or $2.214$ rad