Area of a triangle using two sides and the included angle for HSC Maths Standard 2

Use $A = \frac{1}{2} a b \sin C$ with $a = 14$, $b = 18$, $C = 50\degree$.

$A = \frac{1}{2} \times 14 \times 18 \times \sin 50\degree$.

$\sin 50\degree \approx 0.7660$.

$A = 126 \times 0.7660 \approx 96.52$ cm$^2$.

Markers reward the formula stated, correct substitution, and an answer rounded to two decimal places or sensibly.

$A = \frac{1}{2} \times 4.5 \times 3.8 \times \sin 75\degree$.

$\sin 75\degree \approx 0.9659$.

$A = 0.5 \times 4.5 \times 3.8 \times 0.9659 = 8.55 \times 0.9659 \approx 8.26$ m$^2$.

Rounded to one decimal place: $A \approx 8.3$ m$^2$.

Markers reward the formula, substitution of the included angle (not just any angle), and the final answer at the requested precision.

Choose the formula. Two sides and the angle between them are given, so this is the direct SAS case for $A = \frac{1}{2} a b \sin C$, with $a = 8$, $b = 11$ and the included angle $C = 52\degree$.

Substitute, keeping the calculator in degrees.

Evaluate. Here $\sin 52\degree \approx 0.7880$, so

so the area is $34.67$ m$^2$ to two decimal places. (Check: the answer carries a squared unit, m$^2$, and the $\frac{1}{2}$ was kept; dropping it would have wrongly doubled the area to about $69.3$ m$^2$.)

Identify the parts. The two sides are $a = 9$ and $b = 6$, and the angle between them is $C = 110\degree$, so substitute straight into $A = \frac{1}{2} a b \sin C$.

Substitute and evaluate. The included angle is obtuse, which is fine: $\sin 110\degree \approx 0.9397$ is still positive.

so the area is $25.4$ cm$^2$ to one decimal place. (Check: an obtuse included angle still gives a positive area because $\sin$ is positive between $0\degree$ and $180\degree$; the unit is squared, cm$^2$, as it must be.)

Choose the formula. Two sides and the angle squeezed between them are given, so this is the direct SAS case for $A = \frac{1}{2} a b \sin C$, with $a = 7$, $b = 10$ and the included angle $C = 38\degree$.

Substitute, with the calculator in degrees.

Evaluate. Here $\sin 38\degree \approx 0.6157$, so

so the area is $21.55$ m^2 to two decimal places. (Check: the answer keeps the $\frac{1}{2}$ and carries a squared unit; dropping the $\frac{1}{2}$ would wrongly double the area to about $43.1$ m^2.)

Identify the parts. The two edges are $a = 15$ and $b = 6$, and the angle between them is $C = 40\degree$, so substitute straight into $A = \frac{1}{2} a b \sin C$.

Substitute and evaluate. With $\sin 40\degree \approx 0.6428$,

so the area is $28.9$ m^2 to one decimal place. (Check: the angle used is the one enclosed by the two edges, and the unit is squared, m^2, as it must be.)

Part (a) - area in square metres. The two boundaries and the angle between them give a direct SAS case, with $a = 25$, $b = 32$ and $C = 68\degree$:

Since $\sin 68\degree \approx 0.9272$,

so the area is $370.87$ m$^2$.

Part (b) - convert to hectares. One hectare is $10\,000$ m$^2$, so divide:

so the block is about $0.0371$ ha. (Check: hectares are far larger than square metres, so the number of hectares should be much smaller than the number of square metres, which it is.)

Only three sides are given, so find an angle first. The sides are not a Pythagorean triple ($13^2 + 14^2 = 169 + 196 = 365$, but $15^2 = 225$), so use the cosine rule. Let $A$ be the angle opposite the $13$ m side; it is enclosed by the $14$ m and $15$ m sides:

Now apply the area formula with the two sides enclosing that angle. The sides $14$ m and $15$ m enclose $A$:

so the area is $84$ m$^2$ to the nearest square metre. (Check: the perimeter is $42$ m, comparable to the worked example earlier; the area is plausible for sides this size, and the answer is in m$^2$.)

Find the semi-perimeter $s$. Heron's formula needs $s = \frac{a + b + c}{2}$, so with $a = 9$, $b = 10$ and $c = 17$:

Substitute into Heron's formula. Work out each bracket: $s - a = 9$, $s - b = 8$ and $s - c = 1$.

Evaluate the square root.

so the area is $36.00$ m$^2$. (Check: the cosine rule on the angle between the $9$ m and $10$ m sides gives about $126.87\degree$, and $\frac{1}{2} \times 9 \times 10 \times \sin 126.87\degree \approx 36$ m$^2$, which agrees. Note Heron's formula is not on the Standard 2 reference sheet, so it must be supplied if a question expects it.)

Find the semi-perimeter $s$. Heron's rule needs $s = \frac{a + b + c}{2}$, so with $a = 11$, $b = 13$ and $c = 20$:

Work out each bracket. Subtract each side from $s$: $s - a = 11$, $s - b = 9$ and $s - c = 2$.

Substitute into Heron's rule.

Evaluate the square root.

so the area is $66.0$ m^2 to one decimal place. (Check: the cosine rule on the angle between the $11$ m and $13$ m sides gives about $112.62\degree$, and $\frac{1}{2} \times 11 \times 13 \times \sin 112.62\degree \approx 66$ m^2, which agrees. Heron's rule is not on the Standard 2 reference sheet, so it must be supplied when a question expects it.)

Part (a) - test for a Pythagorean triple. Check whether the two shorter sides satisfy Pythagoras' theorem with the longest side:

Since $60^2 + 80^2 = 100^2$, the triangle is right-angled, with the right angle between the $60$ m and $80$ m sides.

Part (b) - area using the included right angle. With $C = 90\degree$ between the two shorter sides, $\sin 90\degree = 1$, so the area formula collapses to half the product of the perpendicular sides:

so the area is $2400$ m$^2$.

Part (c) - convert to hectares. Divide by $10\,000$:

so the paddock is $0.24$ ha. (Check: spotting the $3$-$4$-$5$ triple scaled by $20$ avoids any cosine-rule work, and the area in m$^2$ is far larger than the area in ha, as expected.)

Three sides only, so find an angle with the cosine rule. The sides are not a Pythagorean triple, so choose the angle $C$ opposite the longest side ($11$ m), which is enclosed by the $6$ m and $9$ m sides:

A negative cosine means the angle is obtuse:

Apply the area formula with the two sides enclosing $C$. Those are the $6$ m and $9$ m sides, and $\sin 92.12\degree \approx 0.9993$:

so the area of the sail is $26.98$ m$^2$. (Check: Heron's formula with $s = \frac{6 + 9 + 11}{2} = 13$ gives $\sqrt{13 \times 7 \times 4 \times 2} = \sqrt{728} \approx 26.98$ m$^2$, which agrees.)

Part (a), area by Heron's rule. Three sides are given and no angle, so use Heron's rule with $a = 40$, $b = 50$ and $c = 60$. First the semi-perimeter:

The brackets are $s - a = 35$, $s - b = 25$ and $s - c = 15$, so

so the area is $992.16$ m^2.

Part (b), number of bags. Each bag covers $200$ m^2, so divide the area by $200$:

You cannot buy part of a bag, and $4$ bags cover only $800$ m^2, which is not enough, so round up to $5$ whole bags.

Part (c), total cost. Five bags at $45 each:

so the seed costs $225. (Check: the cosine rule on the angle between the $40$ m and $50$ m sides gives about $82.82\degree$, and $\frac{1}{2} \times 40 \times 50 \times \sin 82.82\degree \approx 992.16$ m^2, which agrees; $5$ bags cover $1000$ m^2 $> 992.16$ m^2 while $4$ cover only $800$ m^2, so rounding up is correct.)

Part (a) - area from two sides and the included angle. This is a direct SAS case with $a = 18$, $b = 24$ and $C = 70\degree$:

Since $\sin 70\degree \approx 0.9397$,

so the area is $202.97$ m$^2$.

Part (b) - number of bags. Each bag covers $50$ m$^2$, so the number of bags is the area divided by $50$:

You cannot buy part of a bag, and $4$ bags would cover only $200$ m$^2$, which is not quite enough, so round up to $5$ whole bags.

Part (c) - total cost. Five bags at $35 each:

so the fertiliser costs $175. (Check: $5$ bags cover $250$ m$^2 > 202.97$ m$^2$ while $4$ bags cover only $200$ m$^2$, so rounding up is correct; the area carries a squared unit and the $\frac{1}{2}$ was kept throughout.)