Question

when the angle of elevation of the sun is 62°, a telephone pole that is tilted at an angle of 8° directly away from the sun casts a shadow 20 feet long. determine the length of the pole to the nearest foot. the length of the pole is ft (type an integer.)

Explanation:

Step1: Find the angles of the triangle

The triangle formed has a right angle at B? Wait, no, let's analyze the angles. The pole is tilted 8° away from the sun, so the angle between the pole and the vertical (dashed line) is 8°, so the angle inside the triangle at A: the sum of angles in a triangle is 180°. The angle at the base (sun's elevation) is 62°, the angle between the pole and the vertical is 8°, so the angle at the top (angle A) of the triangle: let's see, the triangle has angles: at the base (sun) 62°, at B: 90° + 8°? Wait, no. Wait, the pole is tilted 8° away from the sun, so the angle between the pole and the perpendicular (to the ground) is 8°, so the angle inside the triangle at B is 90° + 8°? Wait, no, let's draw the triangle. The shadow is 20 ft (horizontal), the pole is length \( b \), the sun's angle of elevation is 62°, so the angle at the ground (between shadow and the line to the top of the pole) is 62°. The pole is tilted 8° away from the sun, so the angle between the pole and the vertical (perpendicular to ground) is 8°, so the angle between the pole and the shadow (horizontal) is 90° + 8° = 98°? Wait, no, maybe using the Law of Sines. Let's find the angles of the triangle. The triangle has sides: shadow (20 ft), pole (let's call it \( b \)), and the line from the end of the shadow to the top of the pole (let's call it \( a \)). The angles: angle at the shadow end (sun) is 62°, angle at the pole's base (B) is 90° + 8°? Wait, no, the pole is tilted 8° away from the sun, so the angle between the pole and the vertical is 8°, so the angle between the pole and the ground (horizontal) is 90° - 8° = 82°? Wait, no, if it's tilted away from the sun, the angle between the pole and the ground is 90° + 8°? Wait, maybe I'm overcomplicating. Let's use the Law of Sines. The triangle: angle at the shadow end (let's call it angle C) is 62°, angle at the top of the pole (angle A) is 180° - 62° - (90° + 8°)? Wait, no, the pole is tilted 8° away from the sun, so the angle between the pole and the vertical is 8°, so the angle between the pole and the shadow (horizontal) is 90° + 8° = 98°? Wait, no, the vertical is 90° to the ground, so if the pole is tilted 8° away from the sun, then the angle between the pole and the vertical is 8°, so the angle between the pole and the ground (horizontal) is 90° - 8° = 82°? Wait, maybe the triangle has angles: angle at C (sun) is 62°, angle at B (base of pole) is 90° + 8° = 98°? No, that can't be, because 62 + 98 = 160, so angle at A would be 20°, but that doesn't make sense. Wait, maybe the angle at A is 180° - 62° - (90° - 8°) = 180 - 62 - 82 = 36°? No, I'm confused. Wait, let's re-express: the pole is tilted 8° away from the sun, so the angle between the pole and the perpendicular (to the ground) is 8°, so the angle between the pole and the shadow (horizontal) is 90° + 8° = 98°? Wait, no, the perpendicular to the ground is vertical, so if the pole is tilted 8° away from the sun, then the angle between the pole and the vertical is 8°, so the angle between the pole and the horizontal (shadow) is 90° + 8° = 98°? Then the triangle has angles: at C (sun) 62°, at B (base of pole) 98°, so angle at A (top of pole) is 180 - 62 - 98 = 20°? Then by Law of Sines: \( \frac{b}{\sin 62°} = \frac{20}{\sin 20°} \). Wait, but that would give \( b = \frac{20 \sin 62°}{\sin 20°} \). Let's calculate that. \( \sin 62° \approx 0.8829 \), \( \sin 20° \approx 0.3420 \), so \( b \approx \frac{20 * 0.8829}{0.3420} \approx \frac{17.658}{0.3420} \approx 51.6 \), which is about 52. But wait, maybe the angle at A is 180 - 62 - (90 - 8) = 180 - 62 - 82 = 36°? Then \( \frac{b}{\sin 62°} = \frac{20}{\sin 36°} \), \( \sin 36° \approx 0.5878 \), so \( b \approx \frac{20 * 0.8829}{0.5878} \approx \frac{17.658}{0.5878} \approx 30 \). No, that's different. Wait, maybe I got the angles wrong. Let's look at the diagram: the diagram shows angle A with 8°, and angle at the base 62°, shadow 20 ft. So the triangle has angle at the base (ground) 62°, angle at A (top) 8°? Wait, no, the diagram has angle A with 8°, and the angle at the ground 62°, and the right angle? Wait, the diagram shows a triangle with a right angle? Wait, the diagram has a right angle at B? Wait, the diagram: the shadow is 20 ft (horizontal), the pole is vertical? No, the pole is tilted 8° away from the sun, so the dashed line is vertical, and the pole is 8° from the dashed line. So the triangle has angle at the ground 62°, angle at A (between pole and the line to the sun) 8°? Wait, no, the sum of angles in a triangle: if angle at ground is 62°, angle at A is 8°, then angle at B is 180 - 62 - 8 = 110°? No, that doesn't make sense. Wait, maybe the triangle is a non-right triangle, and we use Law of Sines. Let's re-express: the pole is length \( b \), shadow is 20 ft (side opposite angle A), angle at A is 8°, angle at the ground is 62°, so angle at B is 180 - 62 - 8 = 110°? No, that's not right. Wait, maybe the angle at the top (angle A) is 180 - 62 - (90 - 8) = 180 - 62 - 82 = 36°? I'm getting confused. Wait, let's check the problem again: "a telephone pole that is tilted at an angle of 8° directly away from the sun casts a shadow 20 feet long". So the pole is tilted 8° away from the sun, so the angle between the pole and the vertical (perpendicular to ground) is 8°, so the angle between the pole and the shadow (horizontal) is 90° + 8° = 98°? Then the triangle has angles: at the shadow end (sun) 62°, at the pole's base (B) 98°, so angle at the top (A) is 180 - 62 - 98 = 20°. Then by Law of Sines: \( \frac{b}{\sin 62°} = \frac{20}{\sin 20°} \). Let's compute that: \( \sin 62° \approx 0.8829 \), \( \sin 20° \approx 0.3420 \), so \( b = \frac{20 * 0.8829}{0.3420} \approx \frac{17.658}{0.3420} \approx 51.6 \), so approximately 52. Wait, but maybe the angle at A is 180 - 62 - (90 - 8) = 36°? No, that would be if the pole is tilted towards the sun, but it's tilted away. Wait, let's use the Law of Sines correctly. Let's define the triangle: let’s call the end of the shadow point C, the base of the pole point B, and the top of the pole point A. So CB = 20 ft (shadow), AB = pole length \( b \), angle at C is 62° (angle of elevation), angle at A: the pole is tilted 8° away from the sun, so the angle between AB and the vertical (dashed line) is 8°, so the angle between AB and CB (horizontal) is 90° + 8° = 98°, so angle at B is 98°, angle at C is 62°, so angle at A is 180 - 62 - 98 = 20°. Then Law of Sines: \( \frac{AB}{\sin 62°} = \frac{CB}{\sin 20°} \), so \( AB = \frac{20 \sin 62°}{\sin 20°} \approx \frac{20 * 0.8829}{0.3420} \approx 51.6 \), so to the nearest foot, 52. Wait, but let's check another way. Alternatively, maybe the triangle is a right triangle? No, because the pole is tilted. Wait, maybe the angle at B is 90° - 8° = 82°, so angle at A is 180 - 62 - 82 = 36°, then \( \frac{b}{\sin 62°} = \frac{20}{\sin 36°} \), \( \sin 36° \approx 0.5878 \), so \( b \approx \frac{20 * 0.8829}{0.5878} \approx 30 \), which is different. So which is correct? Let's think about the tilt: if the pole is tilted 8° away from the sun, then the angle between the pole and the vertical is 8°, so the angle between the pole and the horizontal (shadow) is 90° + 8° = 98°, so the triangle has angle at B (base of pole) 98°, angle at C (sun) 62°, so angle at A (top) 20°, so Law of Sines gives \( b \approx 52 \). Let's verify with Law of Sines: \( \frac{20}{\sin 20°} \approx \frac{20}{0.3420} \approx 58.48 \), \( \frac{b}{\sin 62°} \approx \frac{b}{0.8829} \), so \( b \approx 58.48 * 0.8829 \approx 51.6 \), which is 52. So that's the calculation.

Step2: Calculate using Law of Sines

We have a triangle with:
- Side \( CB = 20 \) ft (shadow)
- Angle at \( C = 62° \)
- Angle at \( A = 180° - 62° - 98° = 20° \) (since angle at \( B = 90° + 8° = 98° \) as the pole is tilted 8° away from the sun)

Using the Law of Sines:
\[ \frac{b}{\sin 62°} = \frac{20}{\sin 20°} \]

Solve for \( b \):
\[ b = \frac{20 \cdot \sin 62°}{\sin 20°} \]

Substitute \( \sin 62° \approx 0.8829 \) and \( \sin 20° \approx 0.3420 \):
\[ b \approx \frac{20 \cdot 0.8829}{0.3420} \approx \frac{17.658}{0.3420} \approx 51.6 \]

Round to the nearest foot: \( b \approx 52 \).

Answer:

52