Question

ready
topic: finding the trigonometric ratios in a right triangle
use the given measures on the triangles to write the indicated trig value.

1. $sin p = $

$cos p = $
$\tan p = $

1. $sin \theta = $

$cos \theta = $
$\tan \theta = $

1. $sin b = $

$cos b = $
$\tan b = $

1. $sin a = $

$cos a = $
$\tan a = $
set
topic: drawing solids of revolution
for each of the following solids, draw the two - dimensional shape that would be revolved about the x - axis to generate it.
5.
need help? visit www.rsg support.org
mathematics vision project
licensed under the creative commons attribution cc by 4.0
mathematicsvisionproject.org
10

Explanation:

Problem 1:

Step 1: Recall trigonometric ratios

In a right triangle, \(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\), \(\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}\), \(\tan\theta=\frac{\text{opposite}}{\text{adjacent}}\)

Step 2: Identify sides for \(\angle P\)

For \(\angle P\) in \(\triangle PQR\) (right - angled at \(R\)):

• Opposite side to \(P\): \(QR = 24\)
• Adjacent side to \(P\): \(PR=10\)
• Hypotenuse: \(PQ = 26\)

Step 3: Calculate \(\sin P\)

\(\sin P=\frac{\text{opposite to }P}{\text{hypotenuse}}=\frac{QR}{PQ}=\frac{24}{26}=\frac{12}{13}\)

Step 4: Calculate \(\cos P\)

\(\cos P=\frac{\text{adjacent to }P}{\text{hypotenuse}}=\frac{PR}{PQ}=\frac{10}{26}=\frac{5}{13}\)

Step 5: Calculate \(\tan P\)

\(\tan P=\frac{\text{opposite to }P}{\text{adjacent to }P}=\frac{QR}{PR}=\frac{24}{10}=\frac{12}{5}\)

Problem 2:

Step 1: Recall trigonometric ratios

\(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\), \(\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}\), \(\tan\theta=\frac{\text{opposite}}{\text{adjacent}}\)

Step 2: Identify sides for \(\angle\theta\)

In the right - triangle (right - angled at \(\beta\)):

• Opposite side to \(\theta\): \(7\)
• Adjacent side to \(\theta\): \(40\)
• Hypotenuse: \(41\)

Step 3: Calculate \(\sin\theta\)

\(\sin\theta=\frac{\text{opposite to }\theta}{\text{hypotenuse}}=\frac{7}{41}\)

Step 4: Calculate \(\cos\theta\)

\(\cos\theta=\frac{\text{adjacent to }\theta}{\text{hypotenuse}}=\frac{40}{41}\)

Step 5: Calculate \(\tan\theta\)

\(\tan\theta=\frac{\text{opposite to }\theta}{\text{adjacent to }\theta}=\frac{7}{40}\)

Problem 3:

Step 1: Recall trigonometric ratios

\(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\), \(\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}\), \(\tan\theta=\frac{\text{opposite}}{\text{adjacent}}\)

Step 2: Identify sides for \(\angle B\)

In \(\triangle ABC\) (right - angled at \(C\)):

• Opposite side to \(B\): \(AC = 12\)
• Adjacent side to \(B\): \(BC = 5\)
• Hypotenuse: \(AB=13\)

Step 3: Calculate \(\sin B\)

\(\sin B=\frac{\text{opposite to }B}{\text{hypotenuse}}=\frac{AC}{AB}=\frac{12}{13}\)

Step 4: Calculate \(\cos B\)

\(\cos B=\frac{\text{adjacent to }B}{\text{hypotenuse}}=\frac{BC}{AB}=\frac{5}{13}\)

Step 5: Calculate \(\tan B\)

\(\tan B=\frac{\text{opposite to }B}{\text{adjacent to }B}=\frac{AC}{BC}=\frac{12}{5}\)

Problem 4:

Answer:

s:

1. \(\sin P=\frac{12}{13}\), \(\cos P=\frac{5}{13}\), \(\tan P=\frac{12}{5}\)
2. \(\sin\theta=\frac{7}{41}\), \(\cos\theta=\frac{40}{41}\), \(\tan\theta=\frac{7}{40}\)
3. \(\sin B=\frac{12}{13}\), \(\cos B=\frac{5}{13}\), \(\tan B=\frac{12}{5}\)
4. \(\sin A=\frac{4}{5}\), \(\cos A=\frac{3}{5}\), \(\tan A=\frac{4}{3}\)
5. The 2 - D shape is a curve (or a closed - curve profile) that matches the left - hand side of the bottle (symmetric about the vertical line through the bottle's axis) and is revolved about the \(x\) - axis. (A rough sketch would show a shape that has the same width - height profile as the bottle, for example, a curve that starts near the \(x\) - axis, narrows, then widens at the bottom, similar to the bottle's silhouette)