Answer:
Proof explained below
Step-by-step explanation:
Tan trigonometric ratio
[tex]\sf \tan(\theta)=\dfrac{O}{A}[/tex]
where:
- [tex]\theta[/tex] is the angle
- O is the side opposite the angle
- A is the side adjacent the angle
Draw a diagram to help visualize the scenario (see attached).
If the angles are complementary, their sum is 90°.
Therefore, let one of the angles be [tex]x[/tex] and the other angle be [tex]90^{\circ}-x[/tex].
Use the tan trig ratio to create two equations with the missing side length.
Right triangle ACD
Given:
- [tex]\theta = x[/tex]
- O = CD (tower)
- A = 9 m
Substituting the given values into the tan trig ratio:
[tex]\implies \tan (x)= \dfrac{\textsf{CD}}{9}[/tex]
Right triangle BCD
Given:
- [tex]\theta = 90^{\circ} - x[/tex]
- O = CD (tower)
- A = 4 m
Substituting the given values into the tan trig ratio:
[tex]\implies \tan (90^{\circ} - x)= \dfrac{\textsf{CD}}{4}[/tex]
[tex]\textsf{As }\:\tan(90^{\circ} - x)= \cot (x):[/tex]
[tex]\implies \cot(x)= \dfrac{\textsf{CD}}{4}[/tex]
[tex]\implies \dfrac{1}{\tan(x)}= \dfrac{\textsf{CD}}{4}[/tex]
[tex]\implies \tan(x)= \dfrac{4}{\textsf{CD}}[/tex]
Therefore, the two equations are:
[tex]\textsf{Equation 1}: \quad \tan (x)= \dfrac{\textsf{CD}}{9}[/tex]
[tex]\textsf{Equation 2}: \quad \tan(x)= \dfrac{4}{\textsf{CD}}[/tex]
Substitute one equation into the other and solve for CD:
[tex]\implies \sf \dfrac{CD}{9}=\dfrac{4}{CD}[/tex]
[tex]\implies \sf CD \cdot CD= 4 \cdot 9[/tex]
[tex]\implies \sf CD^2=36[/tex]
[tex]\implies \sf CD=\sqrt{36}[/tex]
[tex]\implies \sf CD= \pm 6[/tex]
As distance cannot be negative, CD = 6 m only, thus proving that the height of the tower is 6 m.
