Help me anyone ASAP snens

Find sin(a)&cos(B), tan(a)&cot(B), and sec(a)&csc(B).

Reil [10]

Answer:

Part A) $sin(\alpha)=\frac{4}{7},\ cos(\beta)=\frac{4}{7}$

Part B) $tan(\alpha)=\frac{4}{\sqrt{33}},\ tan(\beta)=\frac{4}{\sqrt{33}}$

Part C) $sec(\alpha)=\frac{7}{\sqrt{33}},\ csc(\beta)=\frac{7}{\sqrt{33}}$

Step-by-step explanation:

Part A) Find $sin(\alpha)\ and\ cos(\beta)$

we know that

If two angles are complementary, then the value of sine of one angle is equal to the cosine of the other angle

In this problem

$\alpha+\beta=90^o$ ---> by complementary angles

so

$sin(\alpha)=cos(\beta)$

Find the value of $sin(\alpha)$ in the right triangle of the figure

$sin(\alpha)=\frac{8}{14}$ ---> opposite side divided by the hypotenuse

simplify

$sin(\alpha)=\frac{4}{7}$

therefore

$sin(\alpha)=\frac{4}{7}$

$cos(\beta)=\frac{4}{7}$

Part B) Find $tan(\alpha)\ and\ cot(\beta)$

we know that

If two angles are complementary, then the value of tangent of one angle is equal to the cotangent of the other angle

In this problem

$\alpha+\beta=90^o$ ---> by complementary angles

so

$tan(\alpha)=cot(\beta)$

<em>Find the value of the length side adjacent to the angle alpha</em>

Applying the Pythagorean Theorem

Let

x ----> length side adjacent to angle alpha

$14^2=x^2+8^2\\x^2=14^2-8^2\\x^2=132$

$x=\sqrt{132}\ units$

simplify

$x=2\sqrt{33}\ units$

Find the value of $tan(\alpha)$ in the right triangle of the figure

$tan(\alpha)=\frac{8}{2\sqrt{33}}$ ---> opposite side divided by the adjacent side angle alpha

simplify

$tan(\alpha)=\frac{4}{\sqrt{33}}$

therefore

$tan(\alpha)=\frac{4}{\sqrt{33}}$

$tan(\beta)=\frac{4}{\sqrt{33}}$

Part C) Find $sec(\alpha)\ and\ csc(\beta)$

we know that

If two angles are complementary, then the value of secant of one angle is equal to the cosecant of the other angle

In this problem

$\alpha+\beta=90^o$ ---> by complementary angles

so

$sec(\alpha)=csc(\beta)$

Find the value of $sec(\alpha)$ in the right triangle of the figure

$sec(\alpha)=\frac{1}{cos(\alpha)}$

Find the value of $cos(\alpha)$

$cos(\alpha)=\frac{2\sqrt{33}}{14}$ ---> adjacent side divided by the hypotenuse

simplify

$cos(\alpha)=\frac{\sqrt{33}}{7}$

therefore

$sec(\alpha)=\frac{7}{\sqrt{33}}$

$csc(\beta)=\frac{7}{\sqrt{33}}$

6 0

3 years ago

What is equivelint in ratio

fgiga [73]

A equivelint ratio is when u can multi or divide by both sides

6 0

3 years ago

Whats bigger 0.32 or 1/4??

vredina [299]

Answer:

0.32 is bigger

Step-by-step explanation:

change 1/4 to decimal

can I get brainliest

7 0

3 years ago

Read 2 more answers

Simplify the following expression 65l/13q

natima [27]

Answer:

$\dfrac{5l}{q}$

Step-by-step explanation:

$\dfrac{65l}{13q} = \dfrac{(13\cdot 5)l}{13q} = \dfrac{5l}{q}$

3 0

3 years ago

Can someone help me with this please

hichkok12 [17]

<h3>Answer: 4</h3>

===========================================

Explanation:

For now, ignore the variables. Let's just focus on the coefficients 12 and 16.

The GCF of 12 and 16 is 4 because this is the largest factor they have in common.

12 = 4*3

16 = 4*4

You could create a factor tree to get the prime factorization of each number

12 = 2*2*3

16 = 2*2*2*2

Each factor tree has two copies of '2' leading to the GCF 2*2 = 4.

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Now let's go back to the variables. The term 12n and 16w^3 do not have any variables in common. So the final answer won't have any variables in it. If both terms had say a 'w' in them, then w would somehow be involved in the GCF. But that's not the case here. So we just stick to 4 as our answer.

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Side note: If you had 12n+16w^3 and you wanted to factor, then you would factor out the GCF 4 to get 12n+16w^3 = 4(3n+4w^3). Use the distribution rule to confirm this.

4 0

3 years ago