Trigonometry Question
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In triangle ABC, let point D be the intersection of the altitude (from vertex B to side ¯AC) and side ¯AC such that AD+DC=AC. Let m∠ABD=θ and m∠DBC=α. Let T denote the area of a triangle. Prove that sin(θ+α)=sin(θ)cos(α)+sin(α)cos(θ).
Statement | Reason |
1.θ+α=m∠ABC | 1.Angle Addition Postulate |
2.TΔABC=12sin(m∠ABC)AB BC | 2.SAS formula for the area of a triangle |
3.TΔABC=12sin(θ+α)AB BC | 3.Substitution Property of Equality |
4.TΔABC=12AB BCsin(θ+α) | 4. |
5.TΔABD=12sin(θ)AB BD | 5.SAS formula for the area of a triangle |
6.TΔBDC=12sin(α)BD BC | 6.SAS formula for the area of a triangle |
7.¯BD is an altitude of ΔABC | 7. |
8.¯BD⊥¯AC | 8.Definition of an altitude |
9. | 9.Definition of perpendicular lines |
10.ΔADB, ΔBDC are right triangles | 10. |
11. BD BC=cos(α) | 11. |
12.BD=BCcos(α) | 12.Multiplication Property of Equality |
13. BD AB=cos(θ) | 13. |
14.BD=ABcos(θ) | 14.Multiplication Property of Equality |
15.TΔABD=12sin(θ)AB BCcos(α) | 15.Substitution Property of Equality |
16.TΔABD=12AB BCsin(θ)cos(α) | 16.Commutative Property of Multiplication |
17.TΔBDC=12sin(α)ABcos(θ)BC | 17.Substitution Property of Equality |
18.TΔBDC=12AB BCsin(α)cos(θ) | 18.Commutative Property of Multiplication |
19. | 19.Area of a shape is equal to the sum of the areas of its partitioned shapes |
20.12AB BCsin(θ+α)=12AB BCsin(θ)cos(α) +12AB BCsin(α)cos(θ) | 20. |
21.12AB BCsin(θ+α)= 12AB BC(sin(θ)cos(α)+sin(α)cos(θ)) | 21.Distribution Property of Equality |
22.sin(θ+α)=sin(θ)cos(α)+sin(α)cos(θ) | 22.Division Property of Equality |
Grade 11 Trigonometry CCSS: HSF-TF.C.9
- Ratios of right triangles
- SAS Congruence Theorem
- Pythagorean Theorem
- SAS formula for area of a triangle