Math / Trigonometry

QuestionNote: Triangle may not be drawn to scale. Suppose $\mathrm{a}=9$ and $\mathrm{b}=4$ .
Find an exact value or give at least two decimal places: $\sin (A)=$ $\square$ $\cos (A)=$ $\square$ $\tan (A)=$ $\square$ $\sec (A)=$ $\square$ $\csc (A)=$ $\square$ $\cot (A)=$ $\square$

Studdy Solution

STEP 1

1. We are dealing with a right triangle.
2. Side $a = 9$ is opposite angle $A$ .
3. Side $b = 4$ is adjacent to angle $A$ .
4. We need to find the trigonometric ratios for angle $A$ .

STEP 2

1. Use the Pythagorean theorem to find the hypotenuse $c$ .
2. Calculate $\sin(A)$ , $\cos(A)$ , and $\tan(A)$ using the definitions of sine, cosine, and tangent.
3. Calculate $\sec(A)$ , $\csc(A)$ , and $\cot(A)$ using the reciprocal identities.

STEP 3

Use the Pythagorean theorem to find the hypotenuse $c$ :
$c = \sqrt{a^2 + b^2}$
Substitute the given values:
$c = \sqrt{9^2 + 4^2}$ $c = \sqrt{81 + 16}$ $c = \sqrt{97}$

STEP 4

Calculate $\sin(A)$ :
$\sin(A) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{a}{c} = \frac{9}{\sqrt{97}}$
Rationalize the denominator:
$\sin(A) = \frac{9}{\sqrt{97}} \cdot \frac{\sqrt{97}}{\sqrt{97}} = \frac{9\sqrt{97}}{97}$

STEP 5

Calculate $\cos(A)$ :
$\cos(A) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{b}{c} = \frac{4}{\sqrt{97}}$
Rationalize the denominator:
$\cos(A) = \frac{4}{\sqrt{97}} \cdot \frac{\sqrt{97}}{\sqrt{97}} = \frac{4\sqrt{97}}{97}$

STEP 6

Calculate $\tan(A)$ :
$\tan(A) = \frac{\text{opposite}}{\text{adjacent}} = \frac{a}{b} = \frac{9}{4}$

STEP 7

Calculate $\sec(A)$ :
$\sec(A) = \frac{1}{\cos(A)} = \frac{97}{4\sqrt{97}}$
Rationalize the denominator:
$\sec(A) = \frac{97}{4\sqrt{97}} \cdot \frac{\sqrt{97}}{\sqrt{97}} = \frac{97\sqrt{97}}{388}$

STEP 8

Calculate $\csc(A)$ :
$\csc(A) = \frac{1}{\sin(A)} = \frac{97}{9\sqrt{97}}$
Rationalize the denominator:
$\csc(A) = \frac{97}{9\sqrt{97}} \cdot \frac{\sqrt{97}}{\sqrt{97}} = \frac{97\sqrt{97}}{873}$

STEP 9

Calculate $\cot(A)$ :
$\cot(A) = \frac{1}{\tan(A)} = \frac{4}{9}$
The exact values are:
$\sin(A) = \frac{9\sqrt{97}}{97}, \quad \cos(A) = \frac{4\sqrt{97}}{97}, \quad \tan(A) = \frac{9}{4}$
$\sec(A) = \frac{97\sqrt{97}}{388}, \quad \csc(A) = \frac{97\sqrt{97}}{873}, \quad \cot(A) = \frac{4}{9}$

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QuestionNote: Triangle may not be drawn to scale. Suppose $\mathrm{a}=9$ and $\mathrm{b}=4$ .
Find an exact value or give at least two decimal places: $\sin (A)=$ $\square$ $\cos (A)=$ $\square$ $\tan (A)=$ $\square$ $\sec (A)=$ $\square$ $\csc (A)=$ $\square$ $\cot (A)=$ $\square$

Studdy Solution

STEP 1

1. We are dealing with a right triangle.
2. Side $a = 9$ is opposite angle $A$ .
3. Side $b = 4$ is adjacent to angle $A$ .
4. We need to find the trigonometric ratios for angle $A$ .

STEP 2

1. Use the Pythagorean theorem to find the hypotenuse $c$ .
2. Calculate $\sin(A)$ , $\cos(A)$ , and $\tan(A)$ using the definitions of sine, cosine, and tangent.
3. Calculate $\sec(A)$ , $\csc(A)$ , and $\cot(A)$ using the reciprocal identities.

STEP 3

Use the Pythagorean theorem to find the hypotenuse $c$ :
$c = \sqrt{a^2 + b^2}$
Substitute the given values:
$c = \sqrt{9^2 + 4^2}$ $c = \sqrt{81 + 16}$ $c = \sqrt{97}$

STEP 4

Calculate $\sin(A)$ :
$\sin(A) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{a}{c} = \frac{9}{\sqrt{97}}$
Rationalize the denominator:
$\sin(A) = \frac{9}{\sqrt{97}} \cdot \frac{\sqrt{97}}{\sqrt{97}} = \frac{9\sqrt{97}}{97}$

STEP 5

Calculate $\cos(A)$ :
$\cos(A) = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{b}{c} = \frac{4}{\sqrt{97}}$
Rationalize the denominator:
$\cos(A) = \frac{4}{\sqrt{97}} \cdot \frac{\sqrt{97}}{\sqrt{97}} = \frac{4\sqrt{97}}{97}$

STEP 6

Calculate $\tan(A)$ :
$\tan(A) = \frac{\text{opposite}}{\text{adjacent}} = \frac{a}{b} = \frac{9}{4}$

STEP 7

Calculate $\sec(A)$ :
$\sec(A) = \frac{1}{\cos(A)} = \frac{97}{4\sqrt{97}}$
Rationalize the denominator:
$\sec(A) = \frac{97}{4\sqrt{97}} \cdot \frac{\sqrt{97}}{\sqrt{97}} = \frac{97\sqrt{97}}{388}$

STEP 8

Calculate $\csc(A)$ :
$\csc(A) = \frac{1}{\sin(A)} = \frac{97}{9\sqrt{97}}$
Rationalize the denominator:
$\csc(A) = \frac{97}{9\sqrt{97}} \cdot \frac{\sqrt{97}}{\sqrt{97}} = \frac{97\sqrt{97}}{873}$

STEP 9

Calculate $\cot(A)$ :
$\cot(A) = \frac{1}{\tan(A)} = \frac{4}{9}$
The exact values are:
$\sin(A) = \frac{9\sqrt{97}}{97}, \quad \cos(A) = \frac{4\sqrt{97}}{97}, \quad \tan(A) = \frac{9}{4}$
$\sec(A) = \frac{97\sqrt{97}}{388}, \quad \csc(A) = \frac{97\sqrt{97}}{873}, \quad \cot(A) = \frac{4}{9}$

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Studdy Solution

STEP 1

1. We are dealing with a right triangle.2. Side a=9a = 9a=9 is opposite angle AAA.3. Side b=4b = 4b=4 is adjacent to angle AAA.4. We need to find the trigonometric ratios for angle AAA.

STEP 2

STEP 3

Use the Pythagorean theorem to find the hypotenuse ccc:c=a2+b2 c = \sqrt{a^2 + b^2} c=a2+b2​Substitute the given values:c=92+42 c = \sqrt{9^2 + 4^2} c=92+42​ c=81+16 c = \sqrt{81 + 16} c=81+16​ c=97 c = \sqrt{97} c=97​

STEP 4

STEP 5

STEP 6

Calculate tan⁡(A)\tan(A)tan(A):tan⁡(A)=oppositeadjacent=ab=94 \tan(A) = \frac{\text{opposite}}{\text{adjacent}} = \frac{a}{b} = \frac{9}{4} tan(A)=adjacentopposite​=ba​=49​

STEP 7

STEP 8

STEP 9

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1. We are dealing with a right triangle.
2. Side $a = 9$ is opposite angle $A$ .
3. Side $b = 4$ is adjacent to angle $A$ .
4. We need to find the trigonometric ratios for angle $A$ .

Use the Pythagorean theorem to find the hypotenuse $c$ :
$c = \sqrt{a^2 + b^2}$
Substitute the given values:
$c = \sqrt{9^2 + 4^2}$ $c = \sqrt{81 + 16}$ $c = \sqrt{97}$

Calculate $\tan(A)$ :
$\tan(A) = \frac{\text{opposite}}{\text{adjacent}} = \frac{a}{b} = \frac{9}{4}$