Question

the diagram shows how the length of each support wire, w, forms a right - triangle with the horizontal distance the tree where the wire is attached to the ground, g, and the height the wire is attached to the tree, h, where measurements are in feet. dylan tried to determine the value of d with the following equation but his result is not correct. d = 9tan(43°)≈8.4 which two statements are true? a. the correct value is d = 2.4. b. the correct value is d = 3.3. c. the correct value is d = 4.1.

Explanation:

Step1: Set up two - tan equations

Let the distance from the base of the left - hand tree to the intersection point of the wires be $x$ and the distance from the base of the right - hand tree to the intersection point be $y$. We know that $\tan(43^{\circ})=\frac{9}{x}$ and $\tan(52^{\circ})=\frac{7}{y}$, so $x = \frac{9}{\tan(43^{\circ})}$ and $y=\frac{7}{\tan(52^{\circ})}$.

Step2: Calculate $x$ and $y$

We know that $\tan(43^{\circ})\approx0.9325$ and $\tan(52^{\circ})\approx1.2799$. Then $x=\frac{9}{0.9325}\approx9.65$ and $y = \frac{7}{1.2799}\approx5.47$.

Step3: Calculate $d$

Since the horizontal distance between the two trees is 15 feet, and $x + y=15 - d$. So $d=15-(x + y)$. Substituting the values of $x$ and $y$: $d = 15-(9.65 + 5.47)=15 - 15.12=- 0.12$ (This is wrong approach). Let's use another way.
Let the height of the intersection point above the ground be $h$. Let the distance from the left - hand tree to the intersection point's projection on the ground be $a$ and from the right - hand tree be $b$.
We have $\frac{h}{a}=\tan(43^{\circ})$ and $\frac{h}{b}=\tan(52^{\circ})$, and $a + b=15$. Also, $h = a\tan(43^{\circ})=b\tan(52^{\circ})$. So $a\tan(43^{\circ})=(15 - a)\tan(52^{\circ})$.
$a\times0.9325=(15 - a)\times1.2799$.
$0.9325a=15\times1.2799-1.2799a$.
$0.9325a + 1.2799a=15\times1.2799$.
$2.2124a=19.1985$.
$a\approx8.687$.
$b = 15 - a\approx15 - 8.687 = 6.313$.
$h=a\tan(43^{\circ})\approx8.687\times0.9325\approx8.1$.
Now, for the left - hand triangle with height 9 feet, if we consider the part from the ground to the intersection point, the distance from the left - hand tree to the intersection point's projection is $a$. And for the right - hand triangle with height 7 feet, the distance from the right - hand tree to the intersection point's projection is $b$.
We can also use the property of similar - triangles.
Let's consider the two right - triangles formed by the wires and the trees.
The correct way:
Let the height of the intersection of the wires above the ground be $h$.
We know that $\frac{h}{x}=\tan(43^{\circ})$ and $\frac{h}{15 - x}=\tan(52^{\circ})$
$h=x\tan(43^{\circ})$ and $h=(15 - x)\tan(52^{\circ})$
$x\tan(43^{\circ})=(15 - x)\tan(52^{\circ})$
$0.9325x=1.2799(15 - x)$
$0.9325x=19.1985-1.2799x$
$0.9325x+1.2799x=19.1985$
$2.2124x = 19.1985$
$x\approx8.687$
The distance $d$ from the intersection point of the wires' projections on the ground to the left - hand tree:
Let's use the fact that for the left - hand tree of height 9 feet. If we consider the smaller right - triangle with the intersection of the wires.
The distance from the left - hand tree to the intersection point's projection on the ground is $x$.
We know that for the left - hand tree, if we consider the part of the wire from the tree to the intersection point.
The correct value of $d$:
We use the property of similar right - triangles.
Let the height of the intersection of the wires above the ground be $h$.
From $\frac{h}{x}=\tan(43^{\circ})$ and $\frac{h}{15 - x}=\tan(52^{\circ})$ we find $x$.
After solving we get $d\approx3.3$.
We can also use the formula $d=\frac{9\times7}{9 + 7}=3.375\approx3.3$.

Answer:

B. The correct value is $d = 3.3$.