Trigonometry Question
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For ΔABC, where m∠BAC>90°, let the intersection of the altitude from vertex B and the extension of ¯AC be point D (not labeled). Let θ=m∠BAC and α=m∠BAD. Prove that BC2=AC2+AB2-2AC ABcos(θ).

Statement | Reason |
1.¯BD is an altitude | 1.Given |
2.¯BD⊥¯DC | 2.Definition of an altitude |
3.∠BDC is a right angle | 3.Definition of perpendicular lines |
4.ΔBDA, ΔBDC are right triangles | 4.Definition of right triangles |
5. | 5.Cosine ratio in a right triangle |
6.ABcos(α)=AD | 6.Multiplication Property of Equality |
7. | 7.Sine ratio in a right triangle |
8.ABsin(α)=BD | 8.Multiplication Property of Equality |
9.BC2=BD2+CD2 | 9. |
10. | 10.Segment Addition Postulate |
11.BC2=BD2+(AD+AC)2 | 11.Substitution Property of Equality |
12.BC2=(ABsin(α))2+(ABcos(α)+AC)2 | 12. |
13.BC2=AB2sin2(α)+(ABcos(α)+AC)2 | 13.Distributive Property of Exponents |
14.BC2=AB2sin2(α)+AB2cos2(α)+ 2AB ACcos(α)+AC2 | 14.Distributive Property |
15.BC2=AB2(sin2(α)+cos2(α))+ 2AB ACcos(α)+AC2 | 15.Distributive Property |
16.BC2=AB2+2AB ACcos(α)+AC2 | 16. |
17.BC2=AB2+AC2+2AB ACcos(α) | 17.Commutative Property of Addition |
18.α+θ=180° | 18. |
19.α=180°-θ | 19.Subtraction Property of Equality |
20.BC2=AB2+AC2+2AB ACcos(180°-θ) | 20.Substitution Property of Equality |
21.BC2=AB2+AC2-2AB ACcos(θ) | 21.Trigonometric Identity |
Grade 11 Trigonometry CCSS: HSG-SRT.D.10
- Pythagorean Identity
- Pythagorean Theorem
- Algebra (subtraction)
- Subtraction Property of Equality