Question

what is the height, in feet, of the portion ab of the lamppost?\\(\frac{20}{\tan55^{circ}}\\)\\(20cos55^{circ}\\)\\(20\tan55^{circ}\\)\\(\frac{20}{sin55^{circ}}\\)\question 7 (multiple choice worth 1 points)\which pair of angles has congruent values for the \\(sin x^{circ}\\) and the \\(cos y^{circ}\\)?\\(42^{circ}, 42^{circ}\\)\\(42^{circ}, 48^{circ}\\)\\(42^{circ}, 138^{circ}\\)\\(42^{circ}, 132^{circ}\\)\question 8 (multiple choice worth 1 points)\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?

Explanation:

Step1: Analyze right triangle ABX

We have right triangle $ABX$ where $\angle B = 90^\circ$, base $BX = 20$ ft, $\angle AXB = 55^\circ$. We need height $AB$. Use tangent definition: $\tan(\theta)=\frac{\text{opposite}}{\text{adjacent}}$
$\tan(55^\circ)=\frac{AB}{20}$

Step2: Solve for AB

Rearrange to isolate $AB$:
$AB = 20\tan(55^\circ)$

Step1: Use co-function identity

The identity $\sin(x^\circ)=\cos(y^\circ)$ holds when $x + y = 90^\circ$ (complementary angles).

Step2: Check angle pairs

Calculate sum for each pair:

• $42^\circ+42^\circ=84^\circ

eq90^\circ$

• $42^\circ+48^\circ=90^\circ$
• $42^\circ+138^\circ=180^\circ

eq90^\circ$

• $42^\circ+132^\circ=174^\circ

eq90^\circ$

Step1: Find AC first (right triangle ABC)

In $\triangle ABC$, $\angle B=90^\circ$, $BC=12$ m, $\angle C=30^\circ$. Use cosine:
$\cos(30^\circ)=\frac{BC}{AC} \implies AC=\frac{12}{\cos(30^\circ)}=\frac{12}{\frac{\sqrt{3}}{2}}=8\sqrt{3}$ m

Step2: Find AD (right triangle ADC)

In $\triangle ADC$, $\angle ADC=90^\circ$, $\angle C=15^\circ$, $AC=8\sqrt{3}$ m. Use sine:
$\sin(15^\circ)=\frac{AD}{AC} \implies AD=AC\sin(15^\circ)$
$\sin(15^\circ)=\sin(45^\circ-30^\circ)=\sin45\cos30-\cos45\sin30=\frac{\sqrt{6}-\sqrt{2}}{4}$
$AD=8\sqrt{3}\times\frac{\sqrt{6}-\sqrt{2}}{4}=2\sqrt{3}(\sqrt{6}-\sqrt{2})=2(\sqrt{18}-\sqrt{6})=2(3\sqrt{2}-\sqrt{6})=6\sqrt{2}-2\sqrt{6}\approx2.07$ m

Answer:

$20\tan55^\circ$

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