Question

Can someone help with this section? Please ASAP

Answer 1

The values of the trigonometric functions are

7. $sin\theta = \frac{12}{13}$

$cos\theta = \frac{5}{13}$

$tan\theta = \frac{12}{5}$

$csc\theta =\frac{13}{12}$

$sec\theta =\frac{13}{5}$

$cot\theta =\frac{5}{12}$

8. $sin\theta = \frac{\sqrt{7} }{4}$

$cos\theta = \frac{12}{16}$

$tan\theta =\frac{\sqrt{7} }{3}$

$csc\theta = \frac{4}{\sqrt{7} } \ OR \ \frac{4\sqrt{7}}{7}$

$sec\theta =\frac{16}{12}$

$cot\theta = \frac{3}{\sqrt{7}} \ OR \ \frac{3\sqrt{7}}{7}$

Trigonometric functions

From the question, we are to determine the values of the given trigonometric functions

7.

In the given triangle,

Opposite = 12

Adjacent = 5

Hypotenuse = ?

Hyp² = Opp² + Adj² (Pythagorean theorem)

∴ Hyp² = 12² + 5²

Hyp² = 144 + 25

Hyp² = 169

Hyp =√169

Hyp = 13

Using SOH CAH TOA

$sin\theta = \frac{12}{13}$

$cos\theta = \frac{5}{13}$

$tan\theta = \frac{12}{5}$

$csc\theta =\frac{1}{sin\theta} =\frac{13}{12}$

$sec\theta =\frac{1}{cos\theta} =\frac{13}{5}$

$cot\theta =\frac{1}{tan\theta} =\frac{5}{12}$

8.

Hypotenuse = 16

Adjacent = 12

Opposite = ?

Hyp² = Opp² + Adj² (Pythagorean theorem)

Opp² = Hyp² - Adj²

∴ Opp² = 16² - 12²

Opp² = 256 - 144

Opp² = 112

Opp =√112

Opp = 4√7

Using SOH CAH TOA

$sin\theta = \frac{4\sqrt{7} }{16} = \frac{\sqrt{7} }{4}$

$cos\theta = \frac{12}{16}$

$tan\theta = \frac{4\sqrt{7} }{12}= \frac{\sqrt{7} }{3}$

$csc\theta =\frac{1}{sin\theta} =\frac{16}{4\sqrt{7} } = \frac{4}{\sqrt{7} } \ OR \ \frac{4\sqrt{7}}{7}$

$sec\theta =\frac{1}{cos\theta} =\frac{16}{12}$

$cot\theta =\frac{1}{tan\theta} =\frac{12}{4\sqrt{7}} = \frac{3}{\sqrt{7}} \ OR \ \frac{3\sqrt{7}}{7}$

Hence, the values of the trigonometric functions are

7. $sin\theta = \frac{12}{13}$

$cos\theta = \frac{5}{13}$

$tan\theta = \frac{12}{5}$

$csc\theta =\frac{13}{12}$

$sec\theta =\frac{13}{5}$

$cot\theta =\frac{5}{12}$

8. $sin\theta = \frac{\sqrt{7} }{4}$

$cos\theta = \frac{12}{16}$

$tan\theta =\frac{\sqrt{7} }{3}$

$csc\theta = \frac{4}{\sqrt{7} } \ OR \ \frac{4\sqrt{7}}{7}$

$sec\theta =\frac{16}{12}$

$cot\theta = \frac{3}{\sqrt{7}} \ OR \ \frac{3\sqrt{7}}{7}$

Learn more on Trigonometric functions here: brainly.com/question/12172664

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