Question

What is the measure of angle d1

What is the tangent ratio of angle C2

100 POINTS!!!

Answer 1

`\sf \tan\left(c_2\right)=(OD)/(OC)=(2√(66))/(19)`

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:

• CD = AB = 5cm

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:

`\implies OC^2+OD^2=CD^2`

`\implies 3.8^2+OD^2=5^2`

`\implies 14.44+OD^2=25`

`\implies OD^2=10.56`

`\implies OD^2=(1056)/(100)`

`\implies OD^2=(1056 / 4)/(100 / 4)`

`\implies OD^2=(264)/(25)`

`\implies OD^2=(4 \cdot 66)/(25)`

`\implies OD=\sqrt{(4 \cdot 66)/(25)}`

`\implies OD=(√(4 \cdot 66))/(√(25))`

`\implies OD=(2√(66))/(5)`

`\boxed{\begin{minipage}{6 cm}\underline{Tan trigonometric ratio} \\\\$\sf \tan(\theta)=(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}}`

Given:

• θ = c₂
• O = OD
• A = OC

Substitute the values into the tan ratio:

`\implies \sf \tan\left(c_2\right)=(OD)/(OC)`

`\implies \sf \tan\left(c_2\right)=((2√(66))/(5))/(3.8)`

Rewrite 3.8 as 19/5:

`\implies \sf \tan\left(c_2\right)=((2√(66))/(5))/((19)/(5))`

`\implies \sf \tan\left(c_2\right)=(2√(66))/(19)`