Answer:
[tex]\sf \tan\left(c_2\right)=\dfrac{OD}{OC}=\dfrac{2\sqrt{66}}{19}[/tex]
Step-by-step explanation:
**Please note that if the segment AO = 3.8 cm then the 50° angle on the given rhombus is incorrect. It should be 49.46° (2 d.p.)**
In a rhombus, all four sides are equal in length. Therefore:
Diagonals bisect each other at 90°. Therefore:
- OC = AO = 3.8 cm
- m∠COD = 90°
Therefore, triangle COD is a right triangle.
To find the tangent ratio of angle c₂, first find the length of OD using Pythagoras Theorem:
[tex]\implies OC^2+OD^2=CD^2[/tex]
[tex]\implies 3.8^2+OD^2=5^2[/tex]
[tex]\implies 14.44+OD^2=25[/tex]
[tex]\implies OD^2=10.56[/tex]
[tex]\implies OD^2=\dfrac{1056}{100}[/tex]
[tex]\implies OD^2=\dfrac{1056 \div 4}{100 \div 4}[/tex]
[tex]\implies OD^2=\dfrac{264}{25}[/tex]
[tex]\implies OD^2=\dfrac{4 \cdot 66}{25}[/tex]
[tex]\implies OD=\sqrt{\dfrac{4 \cdot 66}{25}}[/tex]
[tex]\implies OD=\dfrac{\sqrt{4 \cdot 66}}{\sqrt{25}}[/tex]
[tex]\implies OD=\dfrac{2\sqrt{66}}{5}[/tex]
[tex]\boxed{\begin{minipage}{6 cm}\underline{Tan trigonometric ratio} \\\\$\sf \tan(\theta)=\dfrac{O}{A}$\\\\where:\\ \phantom{ww}$\bullet$ $\theta$ is the angle. \\ \phantom{ww}$\bullet$ $\sf O$ is the side opposite the angle. \\\phantom{ww}$\bullet$ $\sf A$ is the side adjacent the angle. \\\end{minipage}}[/tex]
Given:
Substitute the values into the tan ratio:
[tex]\implies \sf \tan\left(c_2\right)=\dfrac{OD}{OC}[/tex]
[tex]\implies \sf \tan\left(c_2\right)=\dfrac{\frac{2\sqrt{66}}{5}}{3.8}[/tex]
Rewrite 3.8 as 19/5:
[tex]\implies \sf \tan\left(c_2\right)=\dfrac{\frac{2\sqrt{66}}{5}}{\frac{19}{5}}[/tex]
[tex]\implies \sf \tan\left(c_2\right)=\dfrac{2\sqrt{66}}{19}[/tex]
