1. find side lengths va and an. round to the nearest tenth.

(image of a right triangle with right angle at a, angle at n is 22°, angle at v is 68°, hypotenuse vn is 14)

2. which is greater, an angle measuring 900° or an angle measuring 6π radians? explain.

3. in a right triangle, tanθ = $\frac{7}{24}$. find cosθ. draw a figure to help solve this problem.

4. an angle in standard position has a measure of 7 radians. in which quadrant is the terminal side of the angle?
a) quadrant i
b) quadrant ii
c) quadrant iii
d) quadrant iv

5. brandon wants to purchase stairs for a steep hillside. the length of the staircase is to be 16 feet and needs to end 10 feet higher than the base of the stairs. what is the angle of the hillside? round to the nearest tenth.

Question

1. find side lengths va and an. round to the nearest tenth.

(image of a right triangle with right angle at a, angle at n is 22°, angle at v is 68°, hypotenuse vn is 14)

2. which is greater, an angle measuring 900° or an angle measuring 6π radians? explain.

3. in a right triangle, tanθ = $\frac{7}{24}$. find cosθ. draw a figure to help solve this problem.

4. an angle in standard position has a measure of 7 radians. in which quadrant is the terminal side of the angle?
a) quadrant i
b) quadrant ii
c) quadrant iii
d) quadrant iv

5. brandon wants to purchase stairs for a steep hillside. the length of the staircase is to be 16 feet and needs to end 10 feet higher than the base of the stairs. what is the angle of the hillside? round to the nearest tenth.

Accepted Answer

### Question 1: Find side lengths VA and AN. Round to the nearest tenth.

# Explanation:
## Step1: Identify triangle type and angles
We have a right triangle $ \triangle VAN $ with $ \angle A = 90^\circ $, $ \angle N = 22^\circ $, $ \angle V = 68^\circ $, and hypotenuse $ VN = 14 $.

## Step2: Find $ VA $ (adjacent to $ \angle V $)
Using cosine: $ \cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}} $
For $ \angle V = 68^\circ $, $ VA = VN \cdot \cos(68^\circ) $
$ VA = 14 \cdot \cos(68^\circ) \approx 14 \cdot 0.3746 \approx 5.2 $

## Step3: Find $ AN $ (opposite to $ \angle V $)
Using sine: $ \sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} $
For $ \angle V = 68^\circ $, $ AN = VN \cdot \sin(68^\circ) $
$ AN = 14 \cdot \sin(68^\circ) \approx 14 \cdot 0.9272 \approx 12.98 \approx 13.0 $ (or using $ \angle N = 22^\circ $, $ AN = 14 \cdot \cos(22^\circ) \approx 14 \cdot 0.9272 \approx 13.0 $, $ VA = 14 \cdot \sin(22^\circ) \approx 14 \cdot 0.3746 \approx 5.2 $)

# Answer:
$ VA \approx 5.2 $, $ AN \approx 13.0 $

### Question 2: Which is greater, an angle measuring $ 900^\circ $ or an angle measuring $ 6\pi $ radians? Explain.

# Explanation:
## Step1: Convert $ 900^\circ $ to radians
We know $ 180^\circ = \pi $ radians, so $ 900^\circ = \frac{900}{180} \cdot \pi = 5\pi $ radians.

## Step2: Compare $ 5\pi $ and $ 6\pi $
Since $ 5\pi < 6\pi $, but wait, $ 900^\circ = 5\pi \approx 15.708 $ radians, $ 6\pi \approx 18.8496 $ radians? Wait, no, wait: Wait, $ 360^\circ = 2\pi $ radians, so $ 900^\circ = 900 - 2 \times 360 = 180^\circ $? No, wait, no: $ 900^\circ = 2 \times 360^\circ + 180^\circ $, but in terms of radians, $ 900^\circ = \frac{900}{180} \pi = 5\pi \approx 15.708 $ radians. $ 6\pi \approx 18.8496 $ radians. Wait, but actually, angles can be coterminal, but the question is about the angle measures. Wait, no, $ 900^\circ $ is $ 5\pi $ radians (since $ 180^\circ = \pi $, so $ 900/180 = 5 $), and $ 6\pi $ radians is $ 6 \times 180 = 1080^\circ $. Wait, I made a mistake earlier. $ 180^\circ = \pi $ radians, so to convert degrees to radians: $ \text{radians} = \text{degrees} \times \frac{\pi}{180} $. So $ 900^\circ \times \frac{\pi}{180} = 5\pi \approx 15.708 $ radians. $ 6\pi $ radians is $ 6\pi \approx 18.8496 $ radians. So $ 6\pi $ radians (1080°) is greater than $ 900^\circ $ (5π ≈15.708 radians). Wait, but maybe the question is about the smallest positive coterminal angles? No, the question is about the angle measures. So $ 900^\circ = 5\pi \approx 15.708 $ radians, $ 6\pi \approx 18.8496 $ radians. So $ 6\pi $ radians is greater.

# Answer:
An angle measuring $ 6\pi $ radians is greater. Because $ 900^\circ = 5\pi $ radians (since $ 900^\circ \times \frac{\pi}{180} = 5\pi $), and $ 5\pi < 6\pi $, so $ 6\pi $ radians (≈18.85 radians) is greater than $ 900^\circ $ (≈15.71 radians).

### Question 3: In a right triangle, $ \tan\theta = \frac{7}{24} $. Find $ \cos\theta $. Draw a figure to help solve this problem.

# Explanation:
## Step1: Recall $ \tan\theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{7}{24} $
So, let the opposite side be $ 7k $, adjacent side be $ 24k $ (where $ k > 0 $).

## Step2: Find the hypotenuse $ c $
Using Pythagorean theorem: $ c = \sqrt{(7k)^2 + (24k)^2} = \sqrt{49k^2 + 576k^2} = \sqrt{625k^2} = 25k $

## Step3: Find $ \cos\theta $
$ \cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{24k}{25k} = \frac{24}{25} $ (since $ k
eq 0 $)

# Answer:
$ \cos\theta = \frac{24}{25} $

### Question 4: An angle in standard position has a measure of 7 radians. In which quadrant is the terminal side of the angle?

# Explanation:
## Step1: Recall radian quadrant boundaries
- Quadrant I: $ 0 < \theta < \frac{\pi}{2} \approx 1.5708 $
- Quadrant II: $ \frac{\pi}{2} < \theta < \pi \approx 3.1416 $
- Quadrant III: $ \pi < \theta < \frac{3\pi}{2} \approx 4.7124 $
- Quadrant IV: $ \frac{3\pi}{2} < \theta < 2\pi \approx 6.2832 $
- After $ 2\pi $ (≈6.2832), angles repeat. So $ 7 $ radians: $ 2\pi \approx 6.2832 $, so $ 7 - 2\pi \approx 7 - 6.2832 = 0.7168 $, which is in Quadrant I? Wait, no: Wait, $ 2\pi \approx 6.283 $, $ 3\pi/2 \approx 4.712 $, $ 2\pi \approx 6.283 $, $ 7 $ radians is between $ 2\pi $ (≈6.283) and $ 5\pi/2 $ (≈7.854). Wait, no: Wait, quadrants are every $ \pi/2 $. Let's list the ranges:

- Quadrant I: $ 0 $ to $ \pi/2 $ (≈1.5708)
- Quadrant II: $ \pi/2 $ to $ \pi $ (≈3.1416)
- Quadrant III: $ \pi $ to $ 3\pi/2 $ (≈4.7124)
- Quadrant IV: $ 3\pi/2 $ to $ 2\pi $ (≈6.2832)
- Then, $ 2\pi $ to $ 5\pi/2 $ (≈7.854) is Quadrant I again? Wait, no: Wait, angles in standard position: after $ 2\pi $, it's equivalent to $ \theta - 2\pi $. So $ 7 $ radians: $ 7 - 2\pi \approx 7 - 6.283 = 0.717 $, which is in Quadrant I? But wait, $ 2\pi \approx 6.283 $, $ 7 $ is greater than $ 2\pi $, so subtract $ 2\pi $: $ 7 - 2\pi \approx 0.717 $, which is in Quadrant I? But wait, no, maybe I messed up. Wait, $ \pi \approx 3.14 $, $ 2\pi \approx 6.28 $, $ 3\pi/2 \approx 4.71 $, $ 2\pi \approx 6.28 $, $ 5\pi/2 \approx 7.85 $. So $ 7 $ radians is between $ 2\pi $ (6.28) and $ 5\pi/2 $ (7.85), which is equivalent to $ 7 - 2\pi \approx 0.717 $, which is in Quadrant I? But the options are A) Quadrant I, B) II, C) III, D) IV. Wait, no, wait: Wait, $ 2\pi $ is a full circle, so $ 7 $ radians is $ 2\pi + (7 - 2\pi) $, so the terminal side is the same as $ 7 - 2\pi \approx 0.717 $ radians, which is in Quadrant I? But that seems wrong. Wait, no, wait: Wait, $ \pi \approx 3.14 $, $ 2\pi \approx 6.28 $, $ 3\pi/2 \approx 4.71 $, $ 2\pi \approx 6.28 $, $ 5\pi/2 \approx 7.85 $. So $ 7 $ is between $ 2\pi $ (6.28) and $ 5\pi/2 $ (7.85), which is Quadrant I? But let's check the ranges again. Wait, no: Quadrant I: 0 to $ \pi/2 $ (0 to 1.57), Quadrant II: $ \pi/2 $ to $ \pi $ (1.57 to 3.14), Quadrant III: $ \pi $ to $ 3\pi/2 $ (3.14 to 4.71), Quadrant IV: $ 3\pi/2 $ to $ 2\pi $ (4.71 to 6.28), then Quadrant I again: $ 2\pi $ to $ 5\pi/2 $ (6.28 to 7.85), which is the same as 0 to $ \pi/2 $ (0 to 1.57) because it's a full rotation. So $ 7 - 2\pi \approx 0.717 $, which is in Quadrant I. But the options are A) I, B) II, C) III, D) IV. Wait, but maybe I made a mistake. Wait, $ 2\pi \approx 6.28 $, $ 7 $ is 7, so 7 - 2\pi ≈ 0.717, which is in Quadrant I. So the answer is A? Wait, no, wait: Wait, $ \pi \approx 3.14 $, $ 2\pi \approx 6.28 $, $ 3\pi/2 \approx 4.71 $, $ 2\pi \approx 6.28 $, $ 5\pi/2 \approx 7.85 $. So 7 radians is in Quadrant I? But let's check with another approach. The angle 7 radians: let's see, $ 2\pi \approx 6.28 $, so 7 is 0.72 radians more than $ 2\pi $, so it's in Quadrant I. So the answer is A) Quadrant I? Wait, but maybe the question is considering the angle without subtracting $ 2\pi $, but no, standard position angles are measured from the positive x-axis, counterclockwise. So after $ 2\pi $, it's the same as $ \theta - 2\pi $. So 7 radians is in Quadrant I. But wait, let's check the approximate value: $ \pi \approx 3.14 $, $ 2\pi \approx 6.28 $, $ 3\pi/2 \approx 4.71 $, $ 2\pi \approx 6.28 $, $ 5\pi/2 \approx 7.85 $. So 7 is between $ 2\pi $ (6.28) and $ 5\pi/2 $ (7.85), which is Quadrant I. So the answer is A.

# Answer:
A) Quadrant I

### Question 5: Brandon wants to purchase stairs for a steep hillside. The length of the staircase is to be 16 feet and needs to end 10 feet higher than the base of the stairs. What is the angle of the hillside? Round to the nearest tenth.

# Explanation:
## Step1: Identify triangle components
We have a right triangle where the hypotenuse (staircase length) is 16 feet, and the opposite side (height) is 10 feet. We need to find the angle $ \theta $ (angle of the hillside) using sine: $ \sin\theta = \frac{\text{opposite}}{\text{hypotenuse}} $

## Step2: Calculate $ \theta $
$ \sin\theta = \frac{10}{16} = \frac{5}{8} = 0.625 $
$ \theta = \arcsin(0.625) \approx 38.7^\circ $ (using calculator: $ \arcsin(0.625) \approx 38.68^\circ $, rounded to nearest tenth is $ 38.7^\circ $)

# Answer:
The angle of the hillside is approximately $ 38.7^\circ $

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QUESTION IMAGE

Question

Explanation:

Question 1: Find side lengths VA and AN. Round to the nearest tenth.

Step1: Identify triangle type and angles

Step2: Find \( VA \) (adjacent to \( \angle V \))

Step3: Find \( AN \) (opposite to \( \angle V \))

Step1: Convert \( 900^\circ \) to radians

Step2: Compare \( 5\pi \) and \( 6\pi \)

Step1: Recall \( \tan\theta = \frac{\text{opposite}}{\text{adjacent}} = \frac{7}{24} \)

Step2: Find the hypotenuse \( c \)

Step3: Find \( \cos\theta \)

Answer:

Question 2: Which is greater, an angle measuring \( 900^\circ \) or an angle measuring \( 6\pi \) radians? Explain.