Question

1. in each quadrilateral, calculate the length of gh to the nearest tenth of a centimetre.

a)
b)
c)

Explanation:

Step1: Use tangent function in right - triangle

In right - triangle AGD, $\tan46^{\circ}=\frac{AD}{AG}$, assume $AD = 4.5$ cm. First, find $AG$. Since $\tan46^{\circ}\approx1.0355$, then $AG=\frac{4.5}{\tan46^{\circ}}\approx4.346$ cm. In right - triangle DGH, $\tan66^{\circ}=\frac{GH}{DG}$. First, find $DG$ using Pythagorean theorem in $\triangle AGD$: $DG=\sqrt{AD^{2}+AG^{2}}$. Then, since $\tan66^{\circ}\approx2.246$, $GH = DG\times\tan66^{\circ}$. But we can also use the fact that in the combined right - triangle formed by $\angle GAD$ and $\angle GDH$, $\tan(46^{\circ}+66^{\circ})=\frac{GH}{4.5}$. However, a more straightforward way is to consider the two right - triangles separately. In right - triangle DGH, if we know the length related to the adjacent side of $\angle GDH$. Let's use the fact that in right - triangle AGD, we can find the length of the side related to the larger right - triangle. In right - triangle DGH, $\tan66^{\circ}=\frac{GH}{x}$ (where $x$ is the adjacent side to $\angle GDH$). First, in right - triangle AGD, $\cos46^{\circ}=\frac{AG}{DG}$, $\sin46^{\circ}=\frac{AD}{DG}$. $AG = 4.5\cot46^{\circ}\approx4.35$ cm, $DG=\frac{4.5}{\sin46^{\circ}}\approx6.24$ cm. In right - triangle DGH, $GH = DG\tan66^{\circ}$. $GH\approx6.24\times2.246\approx14.0$ cm.

Step2: Round to nearest tenth

Round the result of $GH$ to the nearest tenth.

Answer:

$14.0$ cm