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3 years ago

Find the remaining sides and angles. Round your answers to the nearest tenth. Show your work.

Mathematics

1 answer:

zubka84 [21]3 years ago

6 0

Answer: These are all the sides and angles:
a = 3
b = 18.76
c = 19
A = 9.1°
B = 80.9°
C = 90°

Explanation:

As per the usual convention the side c is opposite to the angle C, side a is opposite to angle a, and side b is opposite to angle B.

That means that c is the hypotenuse of the triangle.

Then, a is one leg, and b is the other leg.

Therefore, you can use Pythagorean theorem to find the unknown leg, b:

c² = a² + b² ⇒ b² = c² - a²
⇒ b² = 19² - 3² = 352
⇒ b = √352 = 18.76

Now you can use any trigonometric ratio to find one of the unknowm angles.

I wll use sine ratio to find the angle opposite to a:

sin (A) = a / 19 = 3 / 19

⇒ A = arcsin (3/19) ≈ 9.1°.

Then, the other angle is: B = 180° - 9.1° - 90° = 80.9°

And those are all the sides and angles:
a = 3
b = 18.76
c = 19
A = 9.1°
B = 80.9°
C = 90°

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Given tan theta =9, use trigonometric identities to find the exact value of each of the following:_______

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Answer:

$(a)\ \sec^2(\theta) = 82$

$(b)\ \cot(\theta) = \frac{1}{9}$

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Step-by-step explanation:

Given

$\tan(\theta) = 9$

Required

Solve (a) to (d)

Using tan formula, we have:

$\tan(\theta) = \frac{Opposite}{Adjacent}$

This gives:

$\frac{Opposite}{Adjacent} = 9$

Rewrite as:

$\frac{Opposite}{Adjacent} = \frac{9}{1}$

Using a unit ratio;

$Opposite = 9; Adjacent = 1$

Using Pythagoras theorem, we have:

$Hypotenuse^2 = Opposite^2 + Adjacent^2$

$Hypotenuse^2 = 9^2 + 1^2$

$Hypotenuse^2 = 81 + 1$

$Hypotenuse^2 = 82$

Take square roots of both sides

$Hypotenuse =\sqrt{82}$

So, we have:

$Opposite = 9; Adjacent = 1$

$Hypotenuse =\sqrt{82}$

Solving (a):

$\sec^2(\theta)$

This is calculated as:

$\sec^2(\theta) = (\sec(\theta))^2$

$\sec^2(\theta) = (\frac{1}{\cos(\theta)})^2$

Where:

$\cos(\theta) = \frac{Adjacent}{Hypotenuse}$

$\cos(\theta) = \frac{1}{\sqrt{82}}$

So:

$\sec^2(\theta) = (\frac{1}{\cos(\theta)})^2$

$\sec^2(\theta) = (\frac{1}{\frac{1}{\sqrt{82}}})^2$

$\sec^2(\theta) = (\sqrt{82})^2$

$\sec^2(\theta) = 82$

Solving (b):

$\cot(\theta)$

This is calculated as:

$\cot(\theta) = \frac{1}{\tan(\theta)}$

Where:

$\tan(\theta) = 9$ ---- given

So:

$\cot(\theta) = \frac{1}{\tan(\theta)}$

$\cot(\theta) = \frac{1}{9}$

Solving (c):

$\cot(\frac{\pi}{2} - \theta)$

In trigonometry:

$\cot(\frac{\pi}{2} - \theta) = \tan(\theta)$

Hence:

$\cot(\frac{\pi}{2} - \theta) = 9$

Solving (d):

$\csc^2(\theta)$

This is calculated as:

$\csc^2(\theta) = (\csc(\theta))^2$

$\csc^2(\theta) = (\frac{1}{\sin(\theta)})^2$

Where:

$\sin(\theta) = \frac{Opposite}{Hypotenuse}$

$\sin(\theta) = \frac{9}{\sqrt{82}}$

So:

$\csc^2(\theta) = (\frac{1}{\frac{9}{\sqrt{82}}})^2$

$\csc^2(\theta) = (\frac{\sqrt{82}}{9})^2$

$\csc^2(\theta) = \frac{82}{81}$

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Step-by-step explanation:

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