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2 years ago

12

What is always true about the angles of an isosceles triangle? All three angles are congruent. At least two of the angles are co

ngruent. The vertex angle is a right angle. The vertex angle is obtuse.

1 answer:

iragen [17]2 years ago

8 0

Answer:

at least 2 angles are congruent

Step-by-step explanation:

isoseles triangles have two congruent opposite angles such as 70 70 40

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3 years ago

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

Ludmilka [50]

Answer:

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

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

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

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

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:

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

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

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

4 0

3 years ago