Question

Please I need help on these 2 thank you so much

Answer 1

Given the triangle below,

To find GR which is the adjacent of the right-angled triangle above,

Using SOHCAHTOA

$\begin{gathered} \cos \theta=\frac{Adj}{Hyp} \\ \text{Where} \\ \text{Hyp}=18\sqrt[]{3} \\ \text{Adj}=GR \\ \theta=30^0 \end{gathered}$

Substituting the values into the formula above,

$\begin{gathered} \cos 30^0=\frac{GR}{18\sqrt[]{3}} \\ \text{Where} \\ \cos 30^0=\frac{\sqrt[]{3}}{2} \\ \frac{\sqrt[]{3}}{2}=\frac{GR}{18\sqrt[]{3}} \end{gathered}$

Crossmultiply to find GR,

$\begin{gathered} \sqrt[]{3}\times18\sqrt[]{3}=2\times GR \\ 18\times3=2GR \\ \text{Divide both sides by 2} \\ \frac{2GR}{2}=\frac{18\times3}{2} \\ GR=27\text{units} \end{gathered}$

Hence, the third option is the answer.