Solving triangles · Teacher reference · answers included

← Senior 1 · short mathematical inquiries

Teacher reference: contains answers, conditions and proofs.

09 Solving triangles · g10-09-01 · 5–10 min · TEACHER · ANSWERS

A triangle and three equal ratios

If one side changes scale, what happens to a/sinA?

Predict: Fix A=30°,B=60° and double side a. Predict all other sides and the three ratios.

Compare	Computed observation
30–60 triangle Angle A (degrees) = 30; Angle B (degrees) = 60; Opposite side a = 3	The three ratios are equal; the opposite labels matter.
Double scale Angle A (degrees) = 30; Angle B (degrees) = 60; Opposite side a = 6	The three ratios are equal; the opposite labels matter.
Angles sum to180° Angle A (degrees) = 90; Angle B (degrees) = 90; Opposite side a = 3	Positive sides and three positive angles summing to180° are required.

Conclusion: The sine rule gives a/sinA=b/sinB=c/sinC=2R. Holding angles fixes shape; changing one side scales every side and the common ratio.

Conditions: Angles A,B,C are positive and sum to180°; opposite side labels must match.

Justification: Using altitudes, the area equals ab sinC/2=bc sinA/2=ca sinB/2; dividing gives the equal ratios.

Ask: Which angle lies opposite the side labelled a?

Watch for: Pairing a side with an adjacent angle.

Extension: Explain why the common ratio changes under enlargement even though the angles do not.

Exit: A=30°,B=90°,a=3. Find b.

b=3sin90°/sin30°=6.

Before: Triangle angles, sine and cosine. · Timing: predict 1, compare 3, explain 2, exit 1 minute. Finite observations do not prove a universal claim.

09 Solving triangles · g10-09-02 · 5–10 min · TEACHER · ANSWERS

Pythagoras with a correction term

What changes when the included angle is not90°?

Predict: Keep a=5,b=7 and compare c at C=60°,90°,120°.

Compare	Computed observation
Acute included angle Side a = 5; Side b = 7; Included C (degrees) = 60	The correction term depends on the included angle.
Right angle Side a = 5; Side b = 7; Included C (degrees) = 90	The correction term depends on the included angle.
Degenerate boundary Side a = 5; Side b = 7; Included C (degrees) = 180	Boundary angle: a degenerate figure, not a proper triangle.

Conclusion: The cosine rule is c²=a²+b²−2abcosC. The correction vanishes at90°, is negative for acute C and positive for obtuse C.

Conditions: a,b>0 and0°<C<180° for a nondegenerate triangle; C must be their included angle.

Justification: Place the sides at(0,0),(b,0),(acosC,asinC). Squaring the endpoint difference gives a²+b²−2abcosC.

Ask: What is cos90°, and why does that restore Pythagoras?

Watch for: Using a nonincluded angle in the cosine-rule expression.

Extension: Compare c with a+b as C approaches180°.

Exit: For a=5,b=7,C=60°, find c².

25+49−35=39.

Before: Triangle angles, sine and cosine. · Timing: predict 1, compare 3, explain 2, exit 1 minute. Finite observations do not prove a universal claim.

09 Solving triangles · g10-09-03 · 5–10 min · TEACHER · ANSWERS

The same SSA data can make two triangles

How can two sides and a nonincluded angle give0,1 or2 solutions?

Predict: Fix A=30°,b=10. Predict the number of triangles at a=4,5,6,10.

Compare	Computed observation
Too short:0 Known A (degrees) = 30; Known side b = 10; Opposite side a = 4	0 nondegenerate triangle(s); each candidate needs A+B<180°.
Tangent:1 Known A (degrees) = 30; Known side b = 10; Opposite side a = 5	1 nondegenerate triangle(s); each candidate needs A+B<180°.
Ambiguous:2 Known A (degrees) = 30; Known side b = 10; Opposite side a = 6	2 nondegenerate triangle(s); each candidate needs A+B<180°.
Equal sides:1 Known A (degrees) = 30; Known side b = 10; Opposite side a = 10	1 nondegenerate triangle(s); each candidate needs A+B<180°.

Conclusion: The sine rule fixes sinB=b sinA/a. Both B and180°−B may satisfy it, but each must leave C=180°−A−B positive.

Conditions: All sides and angles must be positive. A sine value above1 means no triangle; a zero remaining angle is excluded.

Justification: For0<sinB<1 there are two candidates in(0°,180°). Testing A+B<180° proves which candidates are actual triangles.

Ask: Which second angle has the same sine, and is there room for C?

Watch for: Keeping only the calculator principal arcsine value, or counting a degenerate second triangle.

Extension: Explain the threshold a=b sinA geometrically as a tangent case.

Exit: With A=30°,b=10, how many triangles exist at a=6?

Two: both candidate B values leave a positive C.

Before: Triangle angles, sine and cosine. · Timing: predict 1, compare 3, explain 2, exit 1 minute. Finite observations do not prove a universal claim.

09 Solving triangles · g10-09-04 · 5–10 min · TEACHER · ANSWERS

The same two sides, different areas

Which included angle gives the greatest area for fixed sides?

Predict: For sides5 and7, compare areas at30°,90°,150°.

Compare	Computed observation
30° Side a = 5; Side b = 7; Included C (degrees) = 30	Area follows the altitude, not only the two side lengths.
90° maximum Side a = 5; Side b = 7; Included C (degrees) = 90	Area follows the altitude, not only the two side lengths.
Zero-area boundary Side a = 5; Side b = 7; Included C (degrees) = 0	Boundary angle: a degenerate figure, not a proper triangle.

Conclusion: Area is ab sinC/2. It is greatest at90°; supplementary included angles have the same area but generally different third sides.

Conditions: a,b>0 and0°<C<180°. Endpoint angles give zero area and a degenerate figure.

Justification: Drop the altitude h=a sinC onto side b, so area=bh/2=ab sinC/2. Since sinC≤1, the bound ab/2 is attained at90°.

Ask: Which altitude changes while both given sides stay fixed?

Watch for: Assuming equal two-side data imply equal area regardless of the included angle.

Extension: Find two different included angles with equal nonzero area.

Exit: Find the greatest area with sides4 and6.

12, attained at a90° included angle.

Before: Triangle angles, sine and cosine. · Timing: predict 1, compare 3, explain 2, exit 1 minute. Finite observations do not prove a universal claim.

09 Solving triangles · g10-09-05 · 5–10 min · TEACHER · ANSWERS

Measure a distance without crossing the river

What can a baseline and two sighting angles determine?

Predict: Use a20m baseline and angles45°,60°. Predict whether the far point is above the baseline and whether its distance exceeds20m.

Compare	Computed observation
Unequal sighting angles Baseline (m) = 20; First angle A (degrees) = 45; Second angle B (degrees) = 60	The sightline AP and perpendicular width are different measurements.
Symmetric sighting Baseline (m) = 20; First angle A (degrees) = 45; Second angle B (degrees) = 45	The sightline AP and perpendicular width are different measurements.
No remaining angle Baseline (m) = 20; First angle A (degrees) = 90; Second angle B (degrees) = 90	Need a positive baseline and A+B<180°, with both angles positive.

Conclusion: A baseline c with base angles A,B fixes C=180°−A−B. The sine rule gives AC=c sinB/sinC; its perpendicular river width is AC sinA.

Conditions: c>0,A>0,B>0,A+B<180°. Assume a straight baseline, a common horizontal plane and accurate sighting angles.

Justification: The constructed triangle obeys the sine rule; the right triangle formed by dropping the perpendicular gives width=AC sinA.

Ask: Which segment is the sightline and which one represents width?

Watch for: Calling the slant sightline the perpendicular river width.

Extension: Try two angles whose sum approaches180° and explain why the inferred distance grows.

Exit: For c=20,A=B=45°, find the perpendicular width.

10 metres; the far point is above the midpoint of the baseline.

Before: Triangle angles, sine and cosine. · Timing: predict 1, compare 3, explain 2, exit 1 minute. Finite observations do not prove a universal claim.