All categories

3 years ago

How many triangles can be constructed with angles measuring 80 degrees, 80 degrees and 40 degrees

Mathematics

2 answers:

svetlana [45]3 years ago

8 0

Answer:

None.

Step-by-step explanation:

We have,

The measures of the three angles of a triangle are 80°, 80° and 40°.

Since, we know,

The sum of the three angles of a triangle is 180°.

So, we get,

80° + 80° + 40° = 200°, which is greater than 180°.

Thus, we cannot make any triangle using the given angle measures.

Hence, no triangle can be constructed.

bagirrra123 [75]3 years ago

7 0

Answer:

Infinite triangles

Step-by-step explanation:

A triangle can have angles which total to 180 degrees

When one triangle ABC is constructed with degrees 80, 80 and 40 respectively, then similar triangles can be formed by transforming ABC with a real scale factor say k

k can take any real values positive, negative, rational, irrational etc.

Hence we can say given a triangle ABC with given angles there can be an infinite number of triangles with the same angles which will be similarl to ABC and carry an equal proportion with corrresponding sides.

You might be interested in

How do solve 1.9x(10.6-2.17)

slava [35]

You should do distributive property.

6 0

3 years ago

Read 2 more answers

Solve this please how do you select a image on brainly >~

Paladinen [302]

Answer:

CLICK THE ADD ATTACHMENT

Step-by-step explanation:

3 0

3 years ago

maya wants to burs 600 calories, so far, she burned 584.3 calories. how many more calories must maya burn

MrMuchimi

Answer:

15.7

Step-by-step explanation:

600 - 584.3 = 15.7

5 0

2 years ago

How to find the vertex calculus 2What is the vertex, focus and directrix of x^2 = 6y

son4ous [18]

Solution:

Given:

$x^2=6y$

Part A:

The vertex of an up-down facing parabola of the form;

$\begin{gathered} y=ax^2+bx+c \\ is \\ x_v=-\frac{b}{2a} \end{gathered}$

Rewriting the equation given;

$\begin{gathered} 6y=x^2 \\ y=\frac{1}{6}x^2 \\ \\ \text{Hence,} \\ a=\frac{1}{6} \\ b=0 \\ c=0 \\ \\ \text{Hence,} \\ x_v=-\frac{b}{2a} \\ x_v=-\frac{0}{2(\frac{1}{6})} \\ x_v=0 \\ \\ _{} \\ \text{Substituting the value of x into y,} \\ y=\frac{1}{6}x^2 \\ y_v=\frac{1}{6}(0^2) \\ y_v=0 \\ \\ \text{Hence, the vertex is;} \\ (x_v,y_v)=(h,k)=(0,0) \end{gathered}$

Therefore, the vertex is (0,0)

Part B:

A parabola is the locus of points such that the distance to a point (the focus) equals the distance to a line (directrix)

Using the standard equation of a parabola;

$\begin{gathered} 4p(y-k)=(x-h)^2 \\ \\ \text{Where;} \\ (h,k)\text{ is the vertex} \\ |p|\text{ is the focal length} \end{gathered}$

Rewriting the equation in standard form,

$\begin{gathered} x^2=6y \\ 6y=x^2 \\ 4(\frac{3}{2})(y-k)=(x-h)^2 \\ \text{putting (h,k)=(0,0)} \\ 4(\frac{3}{2})(y-0)=(x-0)^2 \\ Comparing\text{to the standard form;} \\ p=\frac{3}{2} \end{gathered}$

Since the parabola is symmetric around the y-axis, the focus is a distance p from the center (0,0)

Hence,

$\begin{gathered} Focus\text{ is;} \\ (0,0+p) \\ =(0,0+\frac{3}{2}) \\ =(0,\frac{3}{2}) \end{gathered}$

Therefore, the focus is;

$(0,\frac{3}{2})$

Part C:

A parabola is the locus of points such that the distance to a point (the focus) equals the distance to a line (directrix)

Using the standard equation of a parabola;

$\begin{gathered} 4p(y-k)=(x-h)^2 \\ \\ \text{Where;} \\ (h,k)\text{ is the vertex} \\ |p|\text{ is the focal length} \end{gathered}$

Rewriting the equation in standard form,

Since the parabola is symmetric around the y-axis, the directrix is a line parallel to the x-axis at a distance p from the center (0,0).

Hence,

$\begin{gathered} Directrix\text{ is;} \\ y=0-p \\ y=0-\frac{3}{2} \\ y=-\frac{3}{2} \end{gathered}$

Therefore, the directrix is;

$y=-\frac{3}{2}$

3 0

1 year ago

Please help 30 points will give brainliest

Vinil7 [7]

Answer:

thats correct

Step-by-step explanation:

8 0

3 years ago

Read 2 more answers