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dsp73
4 years ago
12

A triangle has an angle that measures 90°. What type of triangle could it be? (Choose all that apply)

Mathematics
2 answers:
valentinak56 [21]4 years ago
8 0

Answer:

The triangle could be a right triangle.

Step-by-step explanation:

Right triangles have right angles. (90 degrees). All triangles have 3 angles that add up to 180 degrees. So as long as one of the angles is 90 degrees, it's a right triangle.

bixtya [17]4 years ago
6 0

Answer:

Right angle triangle

Step-by-step explanation:

If any one angle of a triangle is 90, then it is a right angle triangle

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3 years ago
Austin goes to a trade show that sells reptiles. There is a $5.00 entrance fee. He visited a booth that sold geckos for $35 each
vivado [14]
The answer is D because 35x3 is 105 + the $5 fee is 110
7 0
3 years ago
In triangle abc what is the value of cos b A 5/13 B 12/13 C 5/12 D 13/12
Mumz [18]

Answer:

\boxed{Option \ B}

Step-by-step explanation:

In the triangle,

Hypotenuse = 13

Opposite = Perpendicular = 5

Adjacent = Base = 12

Now,

Cos B = \frac{Adjacent}{Hypotenuse}

Cos B = 12/13

If the triangle is just like in the attached file!

5 0
3 years ago
In circle A, ∠BAE ≅ ∠DAE.
kirza4 [7]

The image of circle A is missing, so i have attached it;

Answer:

x = 17

Step-by-step explanation:

In Δs EAB and EAD

We are told that;

∠BAE ≅ ∠DAE

AB = AD = AE

Now,

The Two triangles have two corresponding equal sides and the angles between them are equal, thus, we can say that the two triangles are congruent by SAS (Side Angle Side) postulate of congruence

By using the result of congruence, we can say that;

EB ≅ ED

We are given that;

EB = 3x - 24 and ED = x + 10

Thus,

3x - 24 = x + 10

∴ 3 x - 24 = x + 10

Add 24 to both sides to give;

3x = x + 34

Subtract x from both sides to give;

2x = 34

Divide both sides by 2 to give;

x = 17

5 0
4 years ago
Read 2 more answers
Which portion of the unit circle satisfies the trigonometric inequality cos^2theta + sin^2theta is greater than or equal to 1. A
liberstina [14]

Answer:

Only points on the circle satisfy the given inequality.

Step-by-step explanation:

Given: Unit circle

To find: portion of the unit circle which satisfies the trigonometric inequality \sin ^2\theta +\cos ^2\theta \geq 1

Solution:

In the given figure, OA = 1 unit (as radius of the unit circle equal to 1)

\sin \theta = side opposite to \theta/hypotenuse

\cos  \theta = side adjacent to \theta/hypotenuse

\sin \theta =\frac{AB}{AO}\\\sin \theta =\frac{AB}{1}\\AB=\sin \theta

\cos  \theta=\frac{OB}{AO}\\\cos \theta =\frac{OB}{1}\\OB=\cos \theta

So, coordinates of A = \left ( \cos \theta ,\sin \theta  \right )

For any point (x,y) on the unit circle with centre at origin, equation of circle is given by x^2+y^2=1

Put (x,y)=\left ( \cos \theta ,\sin \theta  \right )

\cos ^2\theta +\sin ^2\theta =1

So, (x,y)=\left ( \cos \theta ,\sin \theta  \right ) satisfies the equation x^2+y^2=1

For points  (x,y)=\left ( \cos \theta ,\sin \theta  \right ) inside the circle, \cos ^2\theta +\sin ^2\theta

For points  (x,y)=\left ( \cos \theta ,\sin \theta  \right ) outside the circle, \cos ^2\theta +\sin ^2\theta >1

So, only points on the circle satisfy the given inequality.

4 0
3 years ago
Read 2 more answers
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