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3 years ago

Using the right triangle below find tangent of angle A

Mathematics

2 answers:

mash [69]3 years ago

5 0

Answer:

Step-by-step explanation:

Triangle ABC is a right angle triangle.

From the given right angle triangle,

AB represents the hypotenuse of the right angle triangle.

With m∠A as the reference angle,

AC represents the adjacent side of the right angle triangle.

BC represents the opposite side of the right angle triangle.

To determine Tan A, we would apply the tangent trigonometric ratio which is expressed as

Tan θ, = opposite side/adjacent sine. Therefore,

Tan A = 4√3/4

Tan A = √3

vaieri [72.5K]3 years ago

4 0

Answer:

tan 60 = √3 or tanФ = √3

Step-by-step explanation:

let Angle at A be Ф.

We know that,

tanФ = opposite / adjacent

From figure,

Opposite side = 4√3

Adjacent Side = 4

tanФ = 4√3 / 4

tanФ = √3

Given that Ф = 60

Angle at A,

tan 60 = √3

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4 years ago

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3 years ago

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riadik2000 [5.3K]

Answer:

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Step-by-step explanation:

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2 years ago

A student is getting ready to take an important oral examination and is concerned about the possibility of having an "on" day or

Tamiku [17]

Answer:

The students should request an examination with 5 examiners.

Step-by-step explanation:

Let X denote the event that the student has an “on” day, and let Y denote the

denote the event that he passes the examination. Then,

$P(Y)=P(Y|X)P(X)+P(Y|X^{c})P(X^{c})$

The events ( $Y|X$ ) follows a Binomial distribution with probability of success 0.80 and the events ( $Y|X^{c}$ ) follows a Binomial distribution with probability of success 0.40.

It is provided that the student believes that he is twice as likely to have an off day as he is to have an on day. Then,

$P(X)=2\cdot P(X^{c})$

Then,

$P(X)+P(X^{c})=1$

⇒

$2P(X^{c})+P(X^{c})=1\\\\3P(X^{c})=1\\\\P(X^{c})=\frac{1}{3}$

Then,

$P(X)=1-P(X^{c})\\=1-\frac{1}{3}\\=\frac{2}{3}$

Compute the probability that the students passes if request an examination with 3 examiners as follows:

$P(Y)=P(Y|X)P(X)+P(Y|X^{c})P(X^{c})$

$=[\sum\limits^{3}_{x=2}{{3\choose x}(0.80)^{x}(1-0.80)^{3-x}}]\times\frac{2}{3}+[\sum\limits^{3}_{x=2}{{3\choose x}(0.40)^{3}(1-0.40)^{3-x}}]\times\frac{1}{3}$

$=0.715$

The probability that the students passes if request an examination with 3 examiners is 0.715.

Compute the probability that the students passes if request an examination with 5 examiners as follows:

$P(Y)=P(Y|X)P(X)+P(Y|X^{c})P(X^{c})$

$=[\sum\limits^{5}_{x=3}{{5\choose x}(0.80)^{x}(1-0.80)^{5-x}}]\times\frac{2}{3}+[\sum\limits^{5}_{x=3}{{5\choose x}(0.40)^{x}(1-0.40)^{5-x}}]\times\frac{1}{3}$

$=0.734$

The probability that the students passes if request an examination with 5 examiners is 0.734.

As the probability of passing is more in case of 5 examiners, the students should request an examination with 5 examiners.

8 0

3 years ago

What does ¿cual? mean in Spanish

masya89 [10]

Its translates which
Hopes this helps

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3 years ago