Respuesta :
Answer:
Question 1)
[tex]\text{ Distance from A to the house is approximately 78.34 meters}\\\text{The height of the house is approximately 50.77 meters}[/tex]
Question 2)
[tex]\text{ Part A) \cos(x)=\frac{12}{13}} \\ \text{ Part B) \sin(90-x)=\frac{12}{13} }[/tex][tex]\text{ Part A) }\\\cos(x)=\frac{12}{13}\\\\\text{ Part B)}\\\sin(90-x)=\frac{12}{13}[/tex]
Step-by-step explanation:
Question 1)
Please first reference below for the image.
Also recall the basic trigonometric functions with SohCahToa.
Part 1)
In order to find the distance, we will need to find the value of [tex]x[/tex]. However, to do so, we must find the value of [tex]c[/tex] first.
First, we will need to find BM. We can use the smaller triangle. BM is opposite of the smaller triangle, and we already know that the hypotenuse is 20 meters. So, we can use sine:
[tex]\sin(A)=\frac{\text{opp}}{\text{hyp}}[/tex]
Substitute 32° for A and 20 for the hypotenuse. Our BM is our opposite, so substitute BM for opposite. This yields:
[tex]\sin(32)=\frac{BM}{20}[/tex]
Solve for BM. Multiply both sides by 20:
[tex]BM=20\sin(32)[/tex]
Don't approximate just yet. In order to make our answer to most precise, we will approximate at the very end.
So, BM is 20sin(32) meters. BM of the smaller triangle is the same as BM of the larger triangle, so BM of the larger triangle is also 20sin(32). Now, we can find the value of [tex]c[/tex]. However, we will first need to find the value of [tex]\angle MBT[/tex]. To do so, notice that [tex]\angle ABM, \angle MBT, \text{ and } 40\textdegree[/tex] form a straight angle. Therefore:
[tex]\angle ABM+\angle MBT+40^\circ=180[/tex]
Simplify:
[tex]\angle ABM+\angle MBT=140[/tex]
So, we will need to find [tex]\angle ABM[/tex], which is [tex]\angle B[/tex] of the smaller triangle. So, we can use the trig functions again. Since we now know BM, we can use cosine. So:
[tex]\cos(B)=\frac{\text{adj}}{\text{opp}}[/tex]
Substitute 20sin(32) for the adjacent and 20 for the opposite. So:
[tex]\cos(B)=\frac{20\sin(32)}{20}[/tex]
To solve for B, take the inverse cosine of both sides. Evaluate:
[tex]B=\cos^{-1}(\frac{20\sin(32)}{20})=58^\circ[/tex]
Therefore, we can substitute 58 for [tex]\angle ABM[/tex] and solve. This yields:
[tex]58+\angle MBT=140\\\Rightarrow\angle MBT=82^\circ[/tex]
So, we can now find c. We know BM and that [tex]\angle B[/tex] (of the larger triangle) is 82°. To find c, then, we will use cosine. So:
[tex]\cos(B)=\frac{\text{adj}}{\text{hyp}}[/tex]
Substitute 82 for B, 20sin(32) for the adjacent BM, and c for the hypotenuse. So:
[tex]\cos(82)=\frac{20\sin(32)}{c}[/tex]
Solve for c by cross-multiplying:
[tex]c=\frac{20\sin(32)}{\cos(82)}[/tex]
Now, finally, we can solve for [tex]x[/tex]. We know [tex]c[/tex] (the hypotenuse) and we want to find [tex]x[/tex]. We also have an angle of 40°. So, we can use cosine:
[tex]\cos(40)=\frac{x}{c}[/tex]
Multiply both sides by [tex]c[/tex]:
[tex]x=c\cos(40)[/tex]
Substitute [tex]c[/tex]:
[tex]x=\frac{20\sin(32)\cos(40)}{\cos(82)}[/tex]
Approximate:
[tex]x\approx58.34\text{ meters}[/tex]
However, this is only [tex]x[/tex]. The distance from the house is [tex]x+20[/tex]. Therefore, the distance from the house is:
[tex]d\approx58.34+20\approx78.34\text{ meters}[/tex]
Part 2)
We want to find the height h. Since we already determined [tex]c[/tex], we can use it to find [tex]h[/tex]. We will use the sine function this time:
[tex]\sin(40)=\frac{h}{c}\\\Rightarrow h=c\sin(40)\\\Rightarrow h=\frac{20\sin(32)\sin(40)}{\cos(82)}\\\Rightarrow h\approx48.95[/tex]
However, we will still need to add the eye level of 182 centimeters or 1.82 meters. So:
[tex]h\approx48.95+1.82\approx50.77\text{ meters}[/tex]
The height of the house is approximately 50.77 meters.
Question 2)
We know that: [tex]\tan(x)=\frac{5}{12}[/tex]
Remember that tangent is the opposite side over the adjacent side. So, let's find the hypotenuse. We can us the Pythagorean Theorem:
[tex]a^2+b^2=c^2[/tex]
Substitute 5 for a and 12 for b. Solve for c:
[tex]c^2=(5)^2+(12)^2\\\Rightarrow c^2=25+144\\\Rightarrow c=\sqrt{169}=13[/tex]
So, our hypotenuse is 13, our adjacent is 12, and our opposite is 5. We can now solve for the others:
A)
[tex]\cos(x)=\frac{\text{adj}}{\text{hyp}}\\\Rightarrow\cos(x)=\frac{12}{13}[/tex]
B)
[tex]\sin(90-x)=\cos(x)\\\text{Note: This is a cofunction identity}\\\Rightarrow\sin(90-x)=\frac{12}{13}[/tex]
Note: The document I found said [tex]\sin^2(90-x)[/tex]. Anyways, it will just be: [tex]\sin^2(90-x)=\frac{144}{169}[/tex]
