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3 years ago

Help! Help me pleaseeee!! I will give brainiest!!! Pleasee!!

Mathematics

1 answer:

belka [17]3 years ago

6 0

divide the shape into 2 triangles and 1 rectangle

left triangle :

hypotenuse = 7 in

height = 6 in

base = x

find the base using pythagoras theorem

x = √( 7² - 6² )

x = √13

x = 3.6 in

find it's area

6 × 3.6 ÷ 2 = 10.8

right triangle :

hypotenuse = 9 in

height = 6 in

base = x

find the base using pythagoras theorem

x = √( 9² - 6² )

x = √45

x = 6.7 in

find area

9 × 6.7 ÷ 2 = 30.15 in

middle rectangle :

16 in - left tri. base = base

16 - 3.6 = 12.4

find area

6 × 12.4 = 74.4

sum of the area

10.8 + 30.15 + 74.4 = 115.35

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$\tan\,\theta =\frac{\sqrt{21} }{2}$

$\sec\,\theta = \frac{-5}{2}$

$cosec\,\theta =\frac{-5}{\sqrt{21} }$

$\cot\,\theta =\frac{2}{\sqrt{21} }$

Step-by-step explanation:

$\cos\theta =\frac{-2}{5}$

As both $sin\,\theta$ ,

$\theta$ lies in the third quadrant.

In the third quadrant,

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$\sin\,\theta =-\sqrt{1-\cos^2\,\theta} \\=-\sqrt{1-(\frac{-2}{5})^2 } \\\\=-\sqrt{1-\frac{4}{25} }\\\\=-\sqrt{\frac{25-4}{25} }\\\\=-\frac{\sqrt{21} }{5}$

$\tan\,\theta = \frac{\sin\,\theta}{\cos\,\theta }\\\\=\frac{\frac{-\sqrt{21} }{5} }{\frac{-2}{5} }\\\\=\frac{\sqrt{21} }{2}$

$\sec\,\theta =\frac{1}{\cos\,\theta }\\\\=\frac{1}{\frac{-2}{5} }\\\\=\frac{-5}{2}$

$\ cosec \,\theta = \frac{1}{sin\,\theta }\\\\=\frac{1}{\frac{-\sqrt{21} }{5} }\\\\=\frac{-5}{\sqrt{21} }$

$\cot\,\theta =\frac{1}{\tan\,\theta}\\\\=\frac{1}{\frac{\sqrt{21} }{2} }\\\\=\frac{2}{\sqrt{21} }$

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3 years ago

Please help asap!!!!!!!! Thanks

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