Find the missing length of the triangle. a= ? b=4 c=12 1/3

Mathematics

1 answer:

Llana [10]3 years ago

3 0

Answer:

a = 12.93

Step-by-step explanation:

Convert to decimal: 12 1/3 = 12.3

Pythagoras (a^2 = b^2 + c^2):

4^2 = 16

12.3^2 = 151.29

16+151.29 = 167.29

a = √167.29

a = 12.93 (rounded to two decimal places)

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9. Given the point (6,-8) values of the six trig function.

lozanna [386]

Answer:

Here's what I get.

Step-by-step explanation:

9. (6, -8)

The reference angle θ is in the fourth quadrant.

∆AOB is a right triangle.

OB² = OA² + AB² = 6² + (-8)² = 36 + 64 = 100

OB = √100 = 10

$\sin \theta = \dfrac{-8}{10} = -\dfrac{4}{5}\\\\\cos \theta =\dfrac{6}{10} = \dfrac{3}{5}\\\\\tan \theta = \dfrac{-8}{6} = -\dfrac{4}{3}\\\\\csc \theta = \dfrac{10}{-8} = -\dfrac{5}{4}\\\\\sec \theta = \dfrac{10}{6} = \dfrac{5}{3}\\\\\cot \theta = \dfrac{6}{-8} = -\dfrac{3}{4}$

10. cot θ = -(√3)/2

The reference angle θ is in the second quadrant.

∆AOB is a right triangle.

OB² = OA² + AB² = (-√3)² + (2)² = 3 + 4 = 7

OB = √7

$\sin \theta = \dfrac{2}{\sqrt{7}} = \dfrac{2\sqrt{7}}{7}\\\\\cos \theta = \dfrac{-\sqrt{3}}{\sqrt{7}} = -\dfrac{\sqrt{21}}{7}\\\\\tan \theta = \dfrac{2}{-\sqrt{3}} = -\dfrac{2\sqrt{3}}{3}\\\\\csc \theta = \dfrac{\sqrt{7}}{2} \\\\\sec \theta = \dfrac{\sqrt{7}}{-\sqrt{3}} = -\dfrac{\sqrt{21}}{3}\\\\\cot \theta = -\dfrac{\sqrt{3}}{2}$