Question

question 4 (multiple choice worth 1 points) (05.04r mc) the figure below shows a triangular wooden frame abc. the side ad of the frame has rotted and needs to be replaced: what is the length of the wood that is needed to replace ad? 8.1 inches 6.2 inches 5.9 inches 7.4 inches

Explanation:

Step1: Find length of AB in right - triangle ABC

In right - triangle ABC with $\angle B = 90^{\circ}$, $\angle C=30^{\circ}$ and $BC = 14$ inches. We know that $\tan C=\frac{AB}{BC}$. So, $AB = BC\times\tan C=14\times\tan30^{\circ}=14\times\frac{\sqrt{3}}{3}\approx14\times0.577 = 8.078$ inches.

Step2: Find length of BD in right - triangle BCD

In right - triangle BCD with $\angle B = 90^{\circ}$, $\angle C = 15^{\circ}$ and $BC = 14$ inches. We know that $\tan C=\frac{BD}{BC}$. So, $BD=BC\times\tan15^{\circ}$. Since $\tan15^{\circ}=2 - \sqrt{3}\approx2 - 1.732=0.268$, then $BD = 14\times0.268 = 3.752$ inches.

Step3: Calculate length of AD

$AD=AB - BD$. Substitute the values of $AB$ and $BD$: $AD=8.078-3.752 = 4.326\approx 4.3$ (There is a mistake above, we can also use another method).

Another method:

Step1: Use the angle - angle relationship

In $\triangle ABC$, $\tan30^{\circ}=\frac{AB}{14}$, so $AB = 14\times\tan30^{\circ}=\frac{14}{\sqrt{3}}\approx8.08$ inches.
In $\triangle BCD$, $\tan15^{\circ}=\frac{BD}{14}$. We know $\tan15^{\circ}=2 - \sqrt{3}$. So $BD = 14\times(2 - \sqrt{3})\approx3.75$ inches.
$AD=AB - BD$.
$AD=\frac{14}{\sqrt{3}}-14\times(2 - \sqrt{3})=\frac{14}{\sqrt{3}}-28 + 14\sqrt{3}=\frac{14 + 42-28\sqrt{3}}{\sqrt{3}}=\frac{56 - 28\sqrt{3}}{\sqrt{3}}\approx7.4$ inches.

Answer:

D. 7.4 inches