Question

as shown in the figure below, aldo is standing 53 feet from the base of a leaning tree. the length of the tree is 63 feet. the angle of elevation from where aldo is standing to the top of the tree is 60°. find the angle, θ, that the tree is leaning. round your answer to the nearest tenth of a degree.

Explanation:

Step1: Apply Law of Sines

In the triangle, we have sides \( a = 53 \), \( b = 63 \), and angle \( A = 60^\circ \). By Law of Sines: \(\frac{\sin B}{63}=\frac{\sin 60^\circ}{53}\)

Step2: Solve for \(\sin B\)

\(\sin B=\frac{63\sin 60^\circ}{53}\approx\frac{63\times0.8660}{53}\approx\frac{54.558}{53}\approx1.029\) (Wait, this can't be. Wait, maybe I mixed up sides. Wait, the tree is length 63, Aldo's distance is 53, angle of elevation 60°. Wait, maybe the triangle is with sides 53, 63, and angle 60° at Aldo. Wait, no, maybe using Law of Cosines or Law of Sines correctly. Wait, let's re-examine. The triangle has sides: Aldo to base: 53, tree length: 63, angle at Aldo: 60°. So we can use Law of Sines: \(\frac{\sin \text{(angle at top)}}{53}=\frac{\sin 60^\circ}{63}\). Wait, no, let's label the triangle: Let’s call Aldo’s position \( A \), base of tree \( B \), top of tree \( C \). So \( AB = 53 \), \( BC = 63 \)? No, \( AC \) is the tree? Wait, no, the tree is leaning, so the length of the tree is \( AC = 63 \), \( AB = 53 \), angle at \( A \) is \( 60^\circ \). We need to find angle \( \theta \) at \( B \) between the vertical and the tree. Wait, maybe first find angle at \( C \), then find \( \theta \). Wait, using Law of Sines: \(\frac{\sin \angle ACB}{53}=\frac{\sin 60^\circ}{63}\)
\(\sin \angle ACB=\frac{53\sin 60^\circ}{63}\approx\frac{53\times0.8660}{63}\approx\frac{45.898}{63}\approx0.7285\)
So \( \angle ACB\approx\arcsin(0.7285)\approx46.8^\circ \)

Step3: Find angle at \( B \)

In triangle, sum of angles is \( 180^\circ \), so angle at \( B = 180^\circ - 60^\circ - 46.8^\circ = 73.2^\circ \)
Then, \( \theta = 90^\circ - 73.2^\circ = 16.8^\circ \)? Wait, no, maybe the vertical angle. Wait, maybe I made a mistake. Wait, let's try again. Wait, the tree is leaning, so the angle between the tree and vertical is \( \theta \). So the triangle: \( AB = 53 \), \( AC = 63 \), angle at \( A = 60^\circ \). Let's find angle at \( B \) (angle between \( AB \) and \( BC \), where \( BC \) is vertical? No, \( BC \) is vertical, and the tree is \( AC = 63 \). So the angle between \( AC \) and \( BC \) is \( \theta \). So in triangle \( ABC \), \( AB = 53 \), \( AC = 63 \), angle at \( A = 60^\circ \). We can find angle at \( C \) first. Wait, Law of Sines: \(\frac{\sin \angle ABC}{63}=\frac{\sin 60^\circ}{53}\)
\(\sin \angle ABC=\frac{63\sin 60^\circ}{53}\approx\frac{63\times0.8660}{53}\approx1.029\) – that's impossible, since sine can't exceed 1. So I must have labeled the triangle wrong. Ah! Wait, the length of the tree is 63, so the side opposite angle \( A \) (60°) is not 63. Wait, maybe the side opposite 60° is the vertical height? No, let's think again. The triangle has sides: Aldo to base (AB) = 53, tree length (AC) = 63, angle at A (angle between AB and AC) = 60°? No, the angle of elevation is from Aldo to top, so angle between AB (horizontal) and AC (line of sight to top) is 60°, but the tree is leaning, so AC is the tree? No, the tree is leaning, so the actual tree length is 63, so the distance from base to top along the tree is 63. So AB = 53 (horizontal), AC = 63 (tree), angle at A = 60° (angle between AB and the line from A to top, which is not the tree? Wait, the figure shows the tree leaning, so the triangle is with AB = 53, BC = 63 (tree), angle at A = 60°. Wait, no, the figure: Aldo at A, base at B, top at C. AB = 53, BC = 63 (tree), angle at A = 60°. Then we can use Law of Cosines to find AC? No, we need angle at B, \( \theta \), between vertical and BC. Wait, maybe the vertical line is from B, so the angle between BC (tree) and vertical is \( \theta \), so the angle between AB (horizontal) and vertical is 90°, so angle between AB and BC is \( 90^\circ - \theta \). Then in triangle ABC, angles sum to 180°: \( 60^\circ + (90^\circ - \theta) + \text{angle at C} = 180^\circ \), so angle at C = \( 30^\circ + \theta \). But maybe better to use Law of Sines correctly. Wait, I think I messed up the sides. Let's correct: Let’s denote: - \( A \): Aldo’s position - \( B \): Base of the tree - \( C \): Top of the tree - \( AB = 53 \) (horizontal distance) - \( AC = 63 \) (length of the tree, leaning) - \( \angle BAC = 60^\circ \) (angle of elevation) We need to find \( \theta \), the angle between the vertical (from B) and the tree (BC? No, AC is the tree). Wait, the vertical line from B is \( BD \), so \( \theta = \angle DBC \). Then \( \angle ABC = 90^\circ + \theta \)? No, if the tree is leaning towards Aldo, then \( \angle ABC = 90^\circ - \theta \). Wait, this is confusing. Let's use Law of Sines properly. In triangle \( ABC \), we have: - Side \( AB = 53 \) - Side \( AC = 63 \) - Angle \( \angle BAC = 60^\circ \) By Law of Sines: \(\frac{\sin \angle ABC}{AC}=\frac{\sin \angle BAC}{BC}\) Wait, no, Law of Sines is \(\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}\), where \( a \) is opposite angle \( A \), etc. So let's define: - Angle \( A = 60^\circ \) (at Aldo) - Side \( a \): opposite angle \( A \), which is \( BC \) (length of tree? No, tree is \( AC = 63 \)). Wait, I think the mistake is in identifying the sides. Let's look at the figure: Aldo is at the left, 53 ft from base (right). The tree is leaning, length 63 ft. Angle of elevation from Aldo to top is 60°. So the triangle is: Aldo (A) ---53ft--- Base (B), Top (C) connected to A (60° angle) and to B (tree length 63ft). So sides: \( AB = 53 \), \( BC = 63 \), \( AC \) is the line of sight? No, the tree is \( BC = 63 \), \( AB = 53 \), angle at A is 60°. So we can use Law of Cosines to find \( AC \), but we need angle at B. Wait, Law of Cosines: \( AC^2 = AB^2 + BC^2 - 2 \cdot AB \cdot BC \cdot \sin(90^\circ - \theta) \)? No, this is getting too convoluted. Wait, maybe the correct approach is: In the triangle with sides 53, 63, and angle 60° between them, we can find the other angles. Wait, using Law of Sines: \(\frac{\sin \theta'}{53}=\frac{\sin 60^\circ}{63}\), where \( \theta' \) is the angle opposite 53, i.e., angle at top. Then \( \sin \theta' = \frac{53 \sin 60^\circ}{63} \approx \frac{53 \times 0.8660}{63} \approx 0.728 \), so \( \theta' \approx 46.8^\circ \). Then the third angle is \( 180 - 60 - 46.8 = 73.2^\circ \). Now, the angle \( \theta \) is the difference between 90° and this third angle? Wait, the third angle is at the base, between the tree and the horizontal. So if the horizontal is 53ft, and the angle at base is 73.2°, then the angle with the vertical is \( 90 - 73.2 = 16.8^\circ \). Yes, that makes sense. So \( \theta \approx 16.8^\circ \). Wait, let's recalculate \( \sin \theta' \): \( \sin \theta' = \frac{53 \times \sin 60^\circ}{63} \approx \frac{53 \times 0.8660254}{63} \approx \frac{45.8993462}{63} \approx 0.72856 \). Then \( \theta' = \arcsin(0.72856) \approx 46.8^\circ \). Then angle at base: \( 180 - 60 - 46.8 = 73.2^\circ \). Then \( \theta = 90 - 73.2 = 16.8^\circ \).

Answer:

\( \boxed{16.8^\circ} \)