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Derive the cosine rule for the side opposite angle A.

Read the idea, work independently, then explain what changed.

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高一必修 第二册(A版).pdf · 6.4 · PDF 45 / printed page 38

Revisit first: Coordinates, bases and divisionDot products and constrained projection

TOPIC 01

Vector proofs in triangles

Turn geometric hypotheses into vector equations; prove midpoint, centroid and orthocenter properties with their assumptions visible.

What you will be able to explain

  • Choose a convenient origin without changing a geometric claim.
  • Derive a median identity from a squared norm.
  • Separate a theorem, its converse and nondegeneracy conditions.

Try it. Leave your reasoning visible.

Use one hint at a time. A correction explains what changed, not just the final answer.

Examples 1–3: study a full demonstration; finish a solution from its first step; then choose a method independently before opening the solution. The remaining exercises may continue after class.

01 / Standard#Worked example

Derive the cosine rule for the side opposite angle A.

AB=u, AC=v, ∣u∣=c, ∣v∣=b, ∣BC∣=aAB=u,\ AC=v,\ |u|=c,\ |v|=b,\ |BC|=a
Skills and prerequisite lessons

Scope: JM02

Working and explanation

BUILD THE REASONING

Hint 1
BC=v−u.
Hint 2
Expand its squared norm.
Worked solution
  1. Use the dot-product expansion.

    a2=∣v−u∣2=b2+c2−2u⋅va^2=|v-u|^2=b^2+c^2-2u\cdot v
  2. Insert the included-angle relation.

    u⋅v=bccos⁡Au\cdot v=bc\cos A

a²=b²+c²−2bc cos A.

Checks and common pitfalls: The angle must be between the two given side vectors.

Reasoning checklist · self / teacher assessment
  • a²=b²+c²−2bc cos A.
  • The angle must be between the two given side vectors.

Think first. Reveal a hint when the class is ready.

Focus on one question

Teacher preparation and assessment

40-minute lesson

MinutesFocusTeaching and student action
5Prediction and diagnosisCompare one numerical triangle with a universal claim.
8Concept derivationDerive midpoint and median identities with A as origin.
10Guided applicationUse dot products to verify right angles and an orthocenter.
12Transfer and explanationProve the rhombus converse and discuss the missing assumption.
5Exit evidenceRepair a one-line proof with an unjustified conclusion.

Question sequence

  • Which origin makes two vertex positions simplest?
  • Where exactly did the parallelogram assumption enter?

Board development

  • Arrange givens, vector translation, identity and conclusion in four lines.
  • For the orthocenter, write one dot-product equation for each altitude and verify the third.

Expected correct and incorrect responses

  • Students say the diagonals are perpendicular because their lengths match. Expand their dot product to reveal the actual side-length condition.
  • Correct: “The parallelogram assumption makes its diagonal vectors u+v and v−u.”
  • Incorrect: “Every quadrilateral with perpendicular diagonals is a rhombus.”

Follow-up questions

  • Can the centroid formula still be evaluated for collinear points, and does it create a nondegenerate triangle?

Remediation

  • Start with the midpoint subtraction E−D before introducing a length formula.

Exit check

  • Write the missing hypothesis in “perpendicular diagonals imply a rhombus”.

Assessment checklist and sample responses

  • Look for a justified origin, correct directed vectors and a return to the geometric claim. Sample: (u+v)·(u−v)=|u|²−|v|², so perpendicular diagonals force equal adjacent sides in a parallelogram.
    Correct: “The parallelogram assumption makes its diagonal vectors u+v and v−u.”

No sign-in. Work stays in this browser. Export before clearing browser data. Written reasoning is assessed with a checklist.

Curriculum and source notes ↗