Checking possible solutions | Year 11

01 / LEARN

Understand the idea.

Test geometric conditions independently

A calculator value must also satisfy geometric conditions. Triangle angles are positive and total $180^{\circ}$ ; larger sides face larger angles. The two shorter sides must together exceed the longest, otherwise the segments cannot close.

For acute-angle SSA data, compare $a$ with the perpendicular height $b sin A$ . Below that height no triangle is possible. In possible cases, calculate angle candidates and test their remaining angles rather than assuming inverse sine finds every solution.

A result must answer the stated problem

Keep intermediate precision, then round the requested result once. Measurements have limited accuracy; extra calculator digits do not add information. Include units, bearing notation or angle restrictions when the problem requires them.

A sine or cosine outside the range from negative one to one signals inconsistent data or a setup error. Recheck the model rather than forcing the ratio into range. Explain why rejected candidates violate the geometry.

02 / WORKED EXAMPLES

See the method in action.

EXAMPLE 01

Worked example 1

A=40°, a=12 cm and b=7 cm. Decide how many triangles fit.

Worked solution

1
Calculate the sine of B.
$sin B = \frac{7 sin 40^{\circ}}{12} \approx 0.374959$
The known opposite pair is twelve centimetres facing forty degrees.
2
Find both sine candidates.
$\begin{aligned} B_{1} & \approx {22.0218}^{\circ} \\ B_{2} & = 180^{\circ} - B_{1} \approx {157.9782}^{\circ} \end{aligned}$
An acute angle and its supplement have equal sines.
3
Reject the supplementary candidate.
$40^{\circ} + B_{2} \approx {197.9782}^{\circ} > 180^{\circ}$
Two angles already exceed the triangle angle sum, leaving no positive third angle.
4
Complete the valid triangle.
$C = 180^{\circ} - 40^{\circ} - B_{1} \approx {117.9782}^{\circ}$
The acute B candidate leaves a positive third angle.
5
State the number and angles of the solutions.
$one triangle: B \approx {22.0}^{\circ}, C \approx {118.0}^{\circ}$
Exactly one of the two possible sine candidates satisfies the angle sum.

Answer

One; B≈22.0°, C≈118.0°.

EXAMPLE 02

Worked example 2

A student claims a triangle has sides 5 m, 7 m and 13 m. Explain why no angle calculation is needed.

Worked solution

1
Identify the longest side.
$13 m$
Triangle inequality compares the longest side with the sum of the other two.
2
Add the two shorter lengths.
$5 + 7 = 12 m$
These are the greatest distance the shorter segments can span when joined end to end.
3
Compare and reject the proposed triangle.
$12 < 13$
The two shorter sides cannot reach across the longest side, so no triangle exists and no angle calculation is needed.

Answer

No triangle exists.

03 / YOUR TURN

Try it yourself.

Check your understanding

Side a=14 faces A=35°. Side b=9 faces B. Is B smaller or larger than A?

Show answer & reasoning

Smaller.

B is smaller than A. Since $b = 9 < 14 = a$ , the side-angle ordering in a triangle gives $B < A = 35^{\circ}$ .

Keep practising

6 worksheet questions · 16 marks

Open worksheet Worked solutions

Read the questions online

Can lengths 7, 9 and 17 form a triangle? Explain.
2 marks
$A = 30^{\circ}$ , $a = 10$ and $b = 14$ . How many triangles fit, and what are the candidate angles B?
3 marks
$A = 50^{\circ}$ , $a = 8$ and $b = 13$ . Explain the absence of a real inverse-sine result.
3 marks
A triangle has $A = 80^{\circ}$ and $B = 40^{\circ}$ . Which of a and b must be longer?
2 marks
A length calculation gives 18.76491 m. Round to one decimal place and explain whether to round each intermediate value too.
3 marks
A proposed solution gives angles $60^{\circ}$ , $70^{\circ}$ and $55^{\circ}$ . State a check that rejects it.
3 marks

Read the worked solutions

Equivalent correct methods earn the corresponding marks. Include units where asked.

Can lengths 7, 9 and 17 form a triangle? Explain.
2 marks
$7 + 9 = 16 < 17$ , violating triangle inequality.
No.
$A = 30^{\circ}$ , $a = 10$ and $b = 14$ . How many triangles fit, and what are the candidate angles B?
3 marks
$sin B = 0.7$ .
Both candidates leave a positive third angle.
Two; $B \approx {44.4}^{\circ}$ or ${135.6}^{\circ}$ .
$A = 50^{\circ}$ , $a = 8$ and $b = 13$ . Explain the absence of a real inverse-sine result.
3 marks
$sin B = \frac{13 sin 50^{\circ}}{8} \approx 1.245 > 1$ .
This is impossible geometry, not insufficient calculator precision.
The data are inconsistent; no triangle exists.
A triangle has $A = 80^{\circ}$ and $B = 40^{\circ}$ . Which of a and b must be longer?
2 marks
Larger angle A faces longer side a.
a
A length calculation gives 18.76491 m. Round to one decimal place and explain whether to round each intermediate value too.
3 marks
The hundredths digit is 6.
Early rounding may change the final result.
18.8 m; keep intermediate precision.
A proposed solution gives angles $60^{\circ}$ , $70^{\circ}$ and $55^{\circ}$ . State a check that rejects it.
3 marks
$60 + 70 + 55 = 185$ .
Recalculate before accepting its side lengths.
Their sum is $185^{\circ}$ , not $180^{\circ}$ .

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Curriculum reference & lesson scope

SACE Board of South Australia

Stage 1 General Mathematics: practical trigonometry, including similarity, right triangles, triangle area, sine/cosine rules and contextual interpretation.

Official curriculum source ↗

In this lesson

Understand the idea.

Test geometric conditions independently

A result must answer the stated problem

See the method in action.

Worked example 1

Calculate the sine of B.

Find both sine candidates.

Reject the supplementary candidate.

Complete the valid triangle.

State the number and angles of the solutions.

Worked example 2

Identify the longest side.

Add the two shorter lengths.

Compare and reject the proposed triangle.

Try it yourself.

Check your understanding

Keep practising