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angle ss1 maths e1622380134992
Figure 1.
e.g e1622381960891
Figure 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, the side AB is Adjacent in Fig. 1 while it is the Opposite side in Fig. 2.

In Figure 1

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 1:

Given the diagram below, write down the value of \( \scriptsize tan \: 2\propto \)
Hence calculate the value of \( \scriptsize x \: and \: \propto \)

Screenshot 2022 05 17 at 19.11.39

Solution:

Consider ΔABC and \( \scriptsize \angle B \)

\( \scriptsize \angle B \: = 2\propto \)

AC = Opposite = 8cm

BC = Adjacent = 5cm

Recall

\( \scriptsize tan \: \theta \: = \normalsize \frac{Opp}{Adj} \)

\( \scriptsize tan \: 2\propto \: = \normalsize \frac{AC}{BC} \\ = \normalsize \frac{8}{5} \\ \scriptsize = 1.6 \)

\( \scriptsize tan \: 2\propto \: = 1.6 \)

\( \scriptsize 2\propto \: = tan^{-1} (1.6) \)

\( \scriptsize 2\propto \: = 57.9946^o\)

\( \scriptsize 2\propto \: \: \approx 58^o\)

\( \scriptsize \propto \: = \normalsize \frac{58^o}{2}\)

\( \scriptsize \propto \: = 28.997\)

\( \scriptsize \propto \: \approx \: 29\)

Consider BCD

Screenshot 2022 05 17 at 19.23.50
\( \scriptsize \propto \: \approx \: 29\)

DC = Opposite = ?

BC = Adjacent = 5cm

\( \scriptsize tan \: \propto \: = \normalsize \frac{Opp}{Adj} \)

\( \scriptsize tan \: \propto \: = \normalsize \frac{\overline{DC}}{5} \)

\( \scriptsize tan \: 28.997 = \normalsize \frac{\overline{DC}}{5} \)

\( \scriptsize \overline{DC} = 5 \: \times \: tan \: 28.997 \)

\( \scriptsize \overline{DC} = 5 \: \times 0.5542\)

\( \scriptsize \overline{DC} = 2.7712 \)

\( \scriptsize \overline{DC} \: \approx \: 2.8 \)

From the Image in the question

\( \scriptsize \overline{AC} \: = \: \overline{AD} \: + \: \overline{DC} \)

\( \scriptsize \overline{AD} \: = \: \overline{AC} \: -\: \overline{DC} \)

\( \scriptsize \overline{AD} \: = \: 8\: -\: 2.7712 \)

\( \scriptsize \overline{AD} \: = 5.23cm\)

Example 2:

A town Y is 200km from town X in a direction 040º. Find the bearing of town Y from town X. How far is Y East of X? (WAEC)

Solution:

ex 2 trig e1622388662449

(a) Bearing of X from Y is 180º + 40º = 220º

(b) Consider right angle triangle ΔXEY

ex2 trig bear e1622389059578

Considering angle Y = 40º

XE = Opposite = ?

XY = Hypotenuse = 200km

Recall

\( \scriptsize Sin \: \theta \: = \normalsize \frac{Opp}{Hyp} \)

\( \scriptsize Sin 40^o = \normalsize \frac{\overline{XE}}{200} \)

i.e \( \scriptsize \overline{XE} = 200 \: \times \: Sin 40^o \)

\( \scriptsize \overline{XE} = 200 \: \times \: 0.6427 \)

\( \scriptsize \overline{XE} = 128.5575 \)

\( \scriptsize \overline{XE} \: \approx \: 128.6 \: km \)

Example 3

The sides of a triangle are in the ratio 4:4:3.

(a) What kind of triangle is it?
(b) Calculate the angles of the triangle to the nearest degree.

Solution:

maths trig e1622390068157

(a) It is an Isosceles Triangle

(b) Opp = 1.5n, Hyp = 4n

Sin θ = \( \frac{opp}{hyp}\)

Sin θ = \( \frac{1.5n}{4n}\)

Sin θ = 0.375

θ = Sin-1 (0.375)

θ = 22.024

θ = 22º ( to the nearest degree)

Therefore, the other angles are \( \frac{180\:-\:22}{2} \\ = \frac{158}{2} \)

= 79º each

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