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SS1: MATHEMATICS - 3RD TERM

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  1. Geometry (Triangles & Polygons) I
    2 Topics
    |
    1 Quiz
  2. Geometry (Triangles & Polygon) II
    2 Topics
    |
    1 Quiz
  3. Geometry (Triangles & Polygon) III
    3 Topics
    |
    1 Quiz
  4. Trigonometry I
    2 Topics
  5. Trigonometry II
    3 Topics
    |
    1 Quiz
  6. Trigonometry III
    3 Topics
    |
    1 Quiz
  7. Mensuration | Plane Shapes
    3 Topics
    |
    1 Quiz
  8. Mensuration | Arcs, Sectors and Segments of Circles
    4 Topics
    |
    1 Quiz
  9. Mensuration | Solid Shapes
    8 Topics
    |
    1 Quiz
  10. Statistics
    2 Topics
    |
    1 Quiz
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Lesson 4, Topic 1
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Solving Right Angle Triangles

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Topic Content:

  • Solving Right Angle Triangles
Screenshot 2024 05 17 at 12.29.42
Fig. 4.1.1
Screenshot 2024 05 17 at 12.38.18
Fig. 4.1.2

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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Dp
Dp
06/03/2022 11:26 AM

Good

D Msgr Egah
01/02/2023 7:09 PM

Great

irene Ahuruonye
15/08/2024 10:22 AM

it is hard at first but its all good 🤓

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