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3 years ago

Plz help me! The perimeter of the triangle shown is 34 yd.

Mathematics

2 answers:

maw [93]3 years ago

7 0

The answer is C just add 7 and a half ad 12 and a third then minus it from 34 the answer is C

brilliants [131]3 years ago

4 0

37/3 ^2 = 1369/9
15/2 ^2 = 225/4
1369/9 + 225/4 = 7501/36
sqaureroot of 7501/36 = 14.43

so closest to B. 14 3/5

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

$\tan \theta = \frac{\sin \theta}{\cos \theta}$

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⇒ $\theta$ lies in the 3rd quadrant.

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$\sec \theta = \frac{1}{-\frac{4}{5}} = -\frac{5}{4}$

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$\sin \theta = \pm \sqrt{1-\cos^2 \theta}$

Substitute the given values we have;

$\sin \theta = \pm\sqrt{1-(\frac{-4}{5})^2 } =\pm\sqrt{1-\frac{16}{25}} =\pm\sqrt{\frac{25-16}{25}} =\pm \sqrt{\frac{9}{25} } = \pm\frac{3}{5}$

Since, sin < 0

⇒ $\sin \theta = -\frac{3}{5}$

now, find $\tan \theta$ :

$\tan \theta = \frac{\sin \theta}{\cos \theta}$

Substitute the given values we have;

$\tan \theta = \frac{-\frac{3}{5} }{-\frac{4}{5} } = \frac{3}{5}\times \frac{5}{4} = \frac{3}{4}$

Therefore, the exact value of:

(a)

$\sec \theta =-\frac{5}{4}$

(b)

$\tan \theta= \frac{3}{4}$

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3 years ago