SS1: MATHEMATICS - 3RD TERM
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Geometry (Triangles & Polygons) I2 Topics|1 Quiz
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Geometry (Triangles & Polygon) II2 Topics|1 Quiz
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Geometry (Triangles & Polygon) III3 Topics|1 Quiz
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Trigonometry I2 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 Shapes3 Topics|1 Quiz
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Mensuration | Arcs, Sectors and Segments of Circles4 Topics|1 Quiz
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Mensuration | Solid Shapes8 Topics|1 Quiz
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Statistics2 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,
B + C = 90º
B = 90º – C
Sin θ = Cos (90 – θ ) = \(\frac{b}{a} \)
And Cos θ = Sin (90 – θ ) = \(\frac{c}{a} \)
Example 4.1.1:
Given the diagram below, write down the value of \( \scriptsize tan \: 2\propto \)
Hence calculate the value
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it is hard at first but its all good 🤓