SS1: MATHEMATICS - 3RD TERM
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Geometry (Triangles & Polygon) I | Week 11 Topic
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Geometry (Triangles & Polygon) II | Week 22 Topics|1 Quiz
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Geometry (Triangles & Polygon) III | Week 3 & 42 Topics|1 Quiz
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Trigonometry I | Week 5 & 62 Topics
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Trigonometry II3 Topics|1 Quiz
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Trigonometry III3 Topics|1 Quiz
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Mensuration (Plane Shapes) | Week 81 Topic|1 Quiz
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Mensuration II4 Topics|1 Quiz
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Mensuration III | Solid Shapes8 Topics|1 Quiz
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Statistics | Week 92 Topics|1 Quiz
Solving Right Angle Triangles
Topic Content:
- Solving Right Angle Triangles


Recall in trigonometric ratios, the Hypotenuse is unique, facing the right angle (90º), however, the Opposite and Adjacent sides depend on the angle (θ) under consideration.
For example, side AB is Adjacent in Fig. 4.1.1 while it is the Opposite side in Fig. 4.1.2.
In Figure 4.1.1 and Figure 4.1.2:
Sin B = \( \frac{Opp}{Hyp} = \frac{b}{a} \)
Cos B = \( \frac{Adj}{Hyp} = \frac{c}{a} \)
Tan B = \( \frac{Opp}{Adj} = \frac{b}{c} \)
Sin C = \( \frac{Opp}{Hyp} = \frac{c}{a} \)
Cos C = \( \frac{Adj}{Hyp} = \frac{b}{a} \)
Tan C = \( \frac{Opp}{Adj} = \frac{c}{b} \)
In ΔABC, B and C are complementary angles (i.e. B + C = 90º)
As seen above, if B = θ, then C = 90 – θ
Therefore,
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it is hard at first but its all good 🤓