Challenge · Sine & Cosine Rule

15 marks

These problems mix the sine rule, cosine rule and the area formula – first DECIDE which to use (opposite pair given? sine rule; otherwise cosine rule). Give answers to 1 d.p. A worked example shows how to start; hints are at the foot. Calculator in degrees.

Worked example: how to start

In triangle $ABC$, $b=12$ cm, $c=15$ cm and the included angle $A=40^\circ$. Work out the area of the triangle.

Two sides and the included angle, so use the area formula.

$\text{Area}=\tfrac12(12)(15)\sin40^\circ$

$=90\sin40^\circ=57.9$ cm$^2$.

1A triangle has sides $b=12$ cm and $c=15$ cm with an included angle $A=40^\circ$. Work out the length of side $a$.[3 marks]

Answer:

2In triangle $ABC$, $A=55^\circ$, $a=10$ cm and $b=8$ cm. Work out the size of the acute angle $B$.[3 marks]

Answer:

3A triangle has two sides of length $11$ cm and $9$ cm with an angle of $70^\circ$ between them. Work out its area.[2 marks]

Answer:

4A triangle has sides $8$ cm, $6$ cm and $5$ cm. Work out the size of its largest angle.[3 marks]

Answer:

5In triangle $ABC$, $B=66^\circ$, $C=48^\circ$ and side $c=AB=9$ cm. Work out the area of triangle $ABC$.[4 marks]

Answer:

★Extension: A ship sails $17$ km due east, then turns and sails $23$ km on a bearing of $118^\circ$. Work out its direct distance from the start, to 1 d.p.

Stuck? Hints (don't peek unless you need to)1. Two sides and the included angle – use the cosine rule.2. A side and its opposite angle are given – use the sine rule for an angle.3. Two sides and the included angle – use $\tfrac12 ab\sin C$.4. The largest angle is opposite the longest side ($8$ cm). Use the cosine rule for an angle.5. Find a third angle and use the sine rule to get another side, then use the area formula.

Solutions & mark scheme · Sine & Cosine Rule

Total: 15 marks

Award the marks shown for each correct step; many of these have more than one valid route, so give method marks for any correct working.

1A triangle has sides $b=12$ cm and $c=15$ cm with an included angle $A=40^\circ$. Work out the length of side $a$.[3]

Model solution

$a^2=12^2+15^2-2(12)(15)\cos40^\circ$

$a^2=369-360\cos40^\circ=93.2\ldots$

$a=9.7$ cm.

Answer: $9.7\text{ cm}$

Marks

✔1Chooses the cosine rule

✔1$a^2=369-360\cos40^\circ$

✔1$a=9.7$ cm

2In triangle $ABC$, $A=55^\circ$, $a=10$ cm and $b=8$ cm. Work out the size of the acute angle $B$.[3]

Model solution

$\dfrac{\sin B}{8}=\dfrac{\sin55^\circ}{10}$

$\sin B=\dfrac{8\sin55^\circ}{10}=0.655\ldots$

$B=40.9^\circ$.

Answer: $40.9^\circ$

Marks

✔1Sine rule for an angle

✔1$\sin B=0.655\ldots$

✔1$B=40.9^\circ$

3A triangle has two sides of length $11$ cm and $9$ cm with an angle of $70^\circ$ between them. Work out its area.[2]

Model solution

$\text{Area}=\tfrac12(11)(9)\sin70^\circ$

$=49.5\sin70^\circ=46.5$ cm$^2$.

Answer: $46.5\text{ cm}^2$

Marks

✔1Uses $\tfrac12 ab\sin C$

✔1$=46.5$ cm$^2$

4A triangle has sides $8$ cm, $6$ cm and $5$ cm. Work out the size of its largest angle.[3]

Model solution

Largest angle $A$ is opposite the $8$ cm side.

$\cos A=\dfrac{6^2+5^2-8^2}{2(6)(5)}=\dfrac{-3}{60}=-0.05$

$A=92.9^\circ$.

Answer: $92.9^\circ$

Marks

✔1Cosine rule for an angle, largest opposite $8$

✔1$\cos A=-0.05$

✔1$A=92.9^\circ$

5In triangle $ABC$, $B=66^\circ$, $C=48^\circ$ and side $c=AB=9$ cm. Work out the area of triangle $ABC$.[4]

Model solution

$A=180^\circ-66^\circ-48^\circ=66^\circ$.

$\dfrac{b}{\sin66^\circ}=\dfrac{9}{\sin48^\circ}$, so $b=\dfrac{9\sin66^\circ}{\sin48^\circ}=11.06$ cm.

Sides $b$ and $c$ meet at $A$: $\text{Area}=\tfrac12\,bc\sin A=\tfrac12(11.06)(9)\sin66^\circ=45.5$ cm$^2$.

Answer: $45.5\text{ cm}^2$

Marks

✔1Third angle $A=66^\circ$

✔1Sine rule for a second side

✔1$b=11.1$ cm

✔1Area $=45.5$ cm$^2$

★Extension

The interior angle at the turn is $180^\circ-(118^\circ-90^\circ)=152^\circ$.

$d^2=17^2+23^2-2(17)(23)\cos152^\circ=289+529-782\cos152^\circ$

$d^2=818+690.5=1508.5$, so $d=38.8$ km.