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1 year ago

What is the first step in constructing an incenter to triangle xyz?

Mathematics

1 answer:

Marizza181 [45]1 year ago

3 0

The first step is constructing an angle bisector for one of the angles of the triangle.

We have to find the first step for the constructing an inscribed circle

<h3>What is the incenter triangle?</h3>

The incenter of a triangle is the point from which the distances to the sides are equal, in this point we can start to construct the inscribed circle in the triangle because the incenter would also be the center of the circumference.

Therefore the first step in constructing an angle bisector for one of the angles of the triangle.

To learn more about the incenter triangle visit:

brainly.com/question/7537975

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You might be interested in

Find the derivative of ln(secx+tanx)

Sliva [168]

If you're using the app, try seeing this answer through your browser: brainly.com/question/3000160

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Find the derivative of

$\mathsf{y=\ell n(sec\,x+tan\,x)}\\\\\\ \mathsf{y=\ell n\!\left(\dfrac{1}{cos\,x}+\dfrac{sin\,x}{cos\,x} \right )}\\\\\\ \mathsf{y=\ell n\!\left(\dfrac{1+sin\,x}{cos\,x} \right )}$

You can treat y as a composite function of x:

$\left\{\! \begin{array}{l} \mathsf{y=\ell n\,u}\\\\ \mathsf{u=\dfrac{1+sin\,x}{cos\,x}} \end{array} \right.$

so use the chain rule to differentiate y:

$\mathsf{\dfrac{dy}{dx}=\dfrac{dy}{du}\cdot \dfrac{du}{dx}}\\\\\\ \mathsf{\dfrac{dy}{dx}=\dfrac{d}{du}(\ell n\,u)\cdot \dfrac{d}{dx}\!\left(\dfrac{1+sin\,x}{cos\,x}\right)}$

The first derivative is 1/u, and the second one can be evaluated by applying the quotient rule:

$\mathsf{\dfrac{dy}{dx}=\dfrac{1}{u}\cdot \dfrac{\frac{d}{dx}(1+sin\,x)\cdot cos\,x-(1+sin\,x)\cdot \frac{d}{dx}(cos\,x)}{(cos\,x)^2}}\\\\\\ \mathsf{\dfrac{dy}{dx}=\dfrac{1}{u}\cdot \dfrac{(0+cos\,x)\cdot cos\,x-(1+sin\,x)\cdot (-\,sin\,x)}{(cos\,x)^2}}$

Multiply out those terms in parentheses:

$\mathsf{\dfrac{dy}{dx}=\dfrac{1}{u}\cdot \dfrac{cos\,x\cdot cos\,x+(sin\,x+sin\,x\cdot sin\,x)}{(cos\,x)^2}}\\\\\\ \mathsf{\dfrac{dy}{dx}=\dfrac{1}{u}\cdot \dfrac{cos^2\,x+sin\,x+sin^2\,x}{(cos\,x)^2}}\\\\\\ \mathsf{\dfrac{dy}{dx}=\dfrac{1}{u}\cdot \dfrac{(cos^2\,x+sin^2\,x)+sin\,x}{(cos\,x)^2}\qquad\quad (but~~cos^2\,x+sin^2\,x=1)}\\\\\\ \mathsf{\dfrac{dy}{dx}=\dfrac{1}{u}\cdot \dfrac{1+sin\,x}{(cos\,x)^2}}$

Substitute back for $\mathsf{u=\dfrac{1+sin\,x}{cos\,x}:}$

$\mathsf{\dfrac{dy}{dx}=\dfrac{1}{~\frac{1+sin\,x}{cos\,x}~}\cdot \dfrac{1+sin\,x}{(cos\,x)^2}}\\\\\\ \mathsf{\dfrac{dy}{dx}=\dfrac{cos\,x}{1+sin\,x}\cdot \dfrac{1+sin\,x}{(cos\,x)^2}}$

Simplifying that product, you get

$\mathsf{\dfrac{dy}{dx}=\dfrac{1}{1+sin\,x}\cdot \dfrac{1+sin\,x}{cos\,x}}\\\\\\ \mathsf{\dfrac{dy}{dx}=\dfrac{1}{cos\,x}}$

∴ $\boxed{\begin{array}{c}\mathsf{\dfrac{dy}{dx}=sec\,x} \end{array}}\quad\longleftarrow\quad\textsf{this is the answer.}$

I hope this helps. =)

Tags: <em>derivative composite function logarithmic logarithm log trigonometric trig secant tangent sec tan chain rule quotient rule differential integral calculus</em>

3 0

3 years ago

Its midpoint i need help

torisob [31]

Answer:

(3.5, 4.8)

Step-by-step explanation:

2.6 - 1.7 = 0.9, 2.6 + 0.9 = 3.5

3 - 1.2 = 1.8, 3 + 1.8 = 4.8

4 0

3 years ago

Read 2 more answers

Linda and Juan went shopping. Linda spent $14 less than Juan.

Rudiy27

Answer: Let x = amount Linda spent.

x=y-$18

Step-by-step explanation:

7 0

3 years ago

What is the measure of EFG in 0 0 below?

const2013 [10]

Answer:

C. 300

Step-by-step explanation:

EFG and EG is the total distance around the circle

EFG + EG = 360 degrees

EFG + 60 = 360

Subtract 60 from each side

EFG +60-60 = 360-60

EFG = 300

5 0

3 years ago

Read 2 more answers

20. The table shows the numberof days you keep a rented movie before returning it and the total cost of renting the

Harlamova29_29 [7]

Step-by-step explanation:

<u>According to given table we have pairs of points:</u>

(4, 6.00), (5, 8.25), (6, 10.50)

<u>The rate of change is:</u>

(8.25 - 6.00)/(5 - 4) = 2.25

(10.50 - 8.25)/(6 - 5) = 2.25

<u>It is 2.25 and the meaning is:</u>

Rental per day is $2.25

5 0

2 years ago