Question

name:date: _____ bell: ___homework 3: trigonometry:ratios & finding missing sides this is a 2-page document! directions: give each trig ratio as a fraction in simplest form.1.- sin $q$ = ___- sin $r$ = ___- cos $q$ = ___- cos $r$ = ___- tan $q$ = ___- tan $r$ = _____directions: solve for $x$. round to the nearest tenth.2.3.4.5.6.7.8.9.

Explanation:

Step1: Find side PQ (right triangle)

First, calculate the length of side PQ using the Pythagorean theorem:
$$PQ = \sqrt{50^2 - 14^2} = \sqrt{2500 - 196} = \sqrt{2304} = 48$$

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For Problem 1 (Trig Ratios):

Step1: Define sides for ∠Q

Opposite=14, Adjacent=48, Hypotenuse=50

• $\sin Q = \frac{14}{50} = \frac{7}{25}$
• $\cos Q = \frac{48}{50} = \frac{24}{25}$
• $\tan Q = \frac{14}{48} = \frac{7}{24}$

Step2: Define sides for ∠R

Opposite=48, Adjacent=14, Hypotenuse=50

• $\sin R = \frac{48}{50} = \frac{24}{25}$
• $\cos R = \frac{14}{50} = \frac{7}{25}$
• $\tan R = \frac{48}{14} = \frac{24}{7}$

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For Problem 2 (Solve for x):

Step1: Use tangent (opposite/adjacent)

$\tan(46^\circ) = \frac{x}{17}$
$x = 17 \times \tan(46^\circ) \approx 17 \times 1.0355 = 17.6$

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For Problem 3 (Solve for x):

Step1: Use sine (opposite/hypotenuse)

$\sin(67^\circ) = \frac{x}{29}$
$x = 29 \times \sin(67^\circ) \approx 29 \times 0.9205 = 26.7$

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For Problem 4 (Solve for x):

Step1: Use cotangent (adjacent/opposite)

$\cot(29^\circ) = \frac{x}{12}$ or $\tan(29^\circ) = \frac{12}{x}$
$x = \frac{12}{\tan(29^\circ)} \approx \frac{12}{0.5543} = 21.6$

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For Problem 5 (Solve for x):

Step1: Use sine (opposite/hypotenuse)

$\sin(16^\circ) = \frac{x}{37}$
$x = 37 \times \sin(16^\circ) \approx 37 \times 0.2756 = 10.2$

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For Problem 6 (Solve for x):

Step1: Use tangent (opposite/adjacent)

$\tan(58^\circ) = \frac{x}{22}$
$x = 22 \times \tan(58^\circ) \approx 22 \times 1.6003 = 35.2$

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For Problem 7 (Solve for x):

Step1: Use tangent (opposite/adjacent)

$\tan(51^\circ) = \frac{x}{15}$
$x = 15 \times \tan(51^\circ) \approx 15 \times 1.2349 = 18.5$

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For Problem 8 (Solve for x):

Step1: Use tangent (opposite/adjacent)

$\tan(37^\circ) = \frac{48}{x}$
$x = \frac{48}{\tan(37^\circ)} \approx \frac{48}{0.7536} = 63.7$

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For Problem 9 (Solve for x):

Step1: Use tangent (opposite/adjacent)

$\tan(24^\circ) = \frac{x}{9}$
$x = 9 \times \tan(24^\circ) \approx 9 \times 0.4452 = 4.0$

Answer:

1.

• $\sin Q = \frac{7}{25}$, $\sin R = \frac{24}{25}$
• $\cos Q = \frac{24}{25}$, $\cos R = \frac{7}{25}$
• $\tan Q = \frac{7}{24}$, $\tan R = \frac{24}{7}$

1. $17.6$
2. $26.7$
3. $21.6$
4. $10.2$
5. $35.2$
6. $18.5$
7. $63.7$
8. $4.0$