Choosing the right rule and multi-step problems

Covers AS91259

A decision guide

First ask: is the triangle right-angled?
- Yes → use SOH CAH TOA (or Pythagoras for a side). These are your Level 1 tools.
- No → choose between the sine and cosine rules.
No right angle — what do you have?
- A side and its opposite angle (a matching pair) → sine rule.
- Two sides and the angle between them, or all three sides → cosine rule.
Need an area? → $\tfrac{1}{2}ab\sin C$ with the included angle (find it first if necessary).

Bearings

A bearing is a direction measured clockwise from North, written with three digits (e.g. $065^\circ$ , $250^\circ$ ).
Draw a North arrow at each point, mark the bearing angle, then find the interior angle of the triangle — usually by subtracting or using angles on a straight line / co-interior angles.

A bearing measured clockwise from a North line at point A round to the direction of B.

Multi-step problems

Real problems often need two triangles or a rule then a rule: e.g. cosine rule to find a side, then sine rule (or the area formula) using that side.
Draw and label a clear diagram first — most method marks come from setting the problem up correctly.
Carry full accuracy through the middle and round only at the very end.

Worked ExampleTwo rules in one problem

A surveyor measures a triangular plot $ABC$ . She finds $AB = 50$ m, $AC = 65$ m, and the angle at $A$ (between them) is $72^\circ$ . Find $BC$ , and then the size of angle $B$ .

Step 1 — Choose the first rule

Two sides and the included angle at $A$ ⇒ cosine rule to find $BC$ (call it $a$ , opposite $A$ ):

$a^2 = 50^2 + 65^2 - 2(50)(65)\cos 72^\circ = 2500 + 4225 - 6500(0.3090)$ $a^2 = 6725 - 2009 = 4716 \Rightarrow BC = \sqrt{4716} \approx 68.7 \text{ m}$

Step 2 — Choose the second rule

Now we have side $a = 68.7$ opposite $A = 72^\circ$ , and want angle $B$ (opposite side $b = AC = 65$ ). A matching pair ⇒ sine rule:

$\frac{\sin B}{65} = \frac{\sin 72^\circ}{68.7} \Rightarrow \sin B = \frac{65 \sin 72^\circ}{68.7} \approx 0.9000$

Step 3 — Inverse sine

$B = \sin^{-1}(0.9000) \approx 64.2^\circ$

Test yourself

Practice by grade

One question each at Achieved, Merit and Excellence. Have a go, then compare with the model answer.

Achieved

A triangle has all three sides known: $8$ , $6$ and $5$ cm. Which rule finds an angle, and why?

Merit

A yacht sails 30 km on a bearing of $040^\circ$ , then turns and sails 20 km on a bearing of $130^\circ$ . Show that the two legs meet at a right angle, and find the direct distance back to the start.

Excellence

From a point $A$ , the angle of elevation to the top of a tower is $32^\circ$ . Walking 25 m closer to $B$ , the angle of elevation becomes $49^\circ$ . Find the height of the tower. Show clear working.

Choosing the right rule and multi-step problems

Covers AS91259

A decision guide

First ask: is the triangle right-angled?
- Yes → use SOH CAH TOA (or Pythagoras for a side). These are your Level 1 tools.
- No → choose between the sine and cosine rules.
No right angle — what do you have?
- A side and its opposite angle (a matching pair) → sine rule.
- Two sides and the angle between them, or all three sides → cosine rule.
Need an area? → $\tfrac{1}{2}ab\sin C$ with the included angle (find it first if necessary).

Bearings

A bearing is a direction measured clockwise from North, written with three digits (e.g. $065^\circ$ , $250^\circ$ ).
Draw a North arrow at each point, mark the bearing angle, then find the interior angle of the triangle — usually by subtracting or using angles on a straight line / co-interior angles.

A bearing measured clockwise from a North line at point A round to the direction of B.

Multi-step problems

Real problems often need two triangles or a rule then a rule: e.g. cosine rule to find a side, then sine rule (or the area formula) using that side.
Draw and label a clear diagram first — most method marks come from setting the problem up correctly.
Carry full accuracy through the middle and round only at the very end.

Worked ExampleTwo rules in one problem

A surveyor measures a triangular plot $ABC$ . She finds $AB = 50$ m, $AC = 65$ m, and the angle at $A$ (between them) is $72^\circ$ . Find $BC$ , and then the size of angle $B$ .

Step 1 — Choose the first rule

Two sides and the included angle at $A$ ⇒ cosine rule to find $BC$ (call it $a$ , opposite $A$ ):

$a^2 = 50^2 + 65^2 - 2(50)(65)\cos 72^\circ = 2500 + 4225 - 6500(0.3090)$ $a^2 = 6725 - 2009 = 4716 \Rightarrow BC = \sqrt{4716} \approx 68.7 \text{ m}$

Step 2 — Choose the second rule

Now we have side $a = 68.7$ opposite $A = 72^\circ$ , and want angle $B$ (opposite side $b = AC = 65$ ). A matching pair ⇒ sine rule:

$\frac{\sin B}{65} = \frac{\sin 72^\circ}{68.7} \Rightarrow \sin B = \frac{65 \sin 72^\circ}{68.7} \approx 0.9000$

Step 3 — Inverse sine

$B = \sin^{-1}(0.9000) \approx 64.2^\circ$

Test yourself

Practice by grade

One question each at Achieved, Merit and Excellence. Have a go, then compare with the model answer.

Achieved

A triangle has all three sides known: $8$ , $6$ and $5$ cm. Which rule finds an angle, and why?

Merit

Excellence