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Soloha48 [4]
3 years ago
14

If PQ and RS intersect to form four right angles which statement is true

Mathematics
2 answers:
Rus_ich [418]3 years ago
7 0
Given that PQ and RS intersect to form four right angles, then they form an angle of 90 degrees, this implies that:
if the meet at point O, 
then;
∠POR=∠POS=∠QOS=∠ROQ=90°
Therefore the correct statement is:
PQ is perpendicular to RS
that is to mean:
PQ⊥RS
The answer is C.
Aliun [14]3 years ago
3 0

Answer:

Given: PQ and RS intersect to form four right angles.

Option C is correct.i.e., PQ ⊥ RS

Because Perpendicular means a line intersecting another at right angles and making 4 right angles between them. Here, also PQ intersect with RS at right angles and makes 4 right angle

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Trigonometry <br> Find x<br> 8.3 <br> 34 degrees
Diano4ka-milaya [45]

Answer:

Step-by-step explanation:

In Triangle

A

B

C

with the right angle at

C

, let

a

,

b

, and

c

be the opposite, the adjacent, and the hypotenuse of

∠

A

. Then, we have

sin

A

=

a

c

⇒

m

∠

A

=

sin

−

1

(

a

c

)

sin

B

=

b

c

⇒

m

∠

B

=

sin

−

1

(

b

c

)

I hope that this was helpful.

Wataru ·  1 · Oct 29 2014

How do you find all the missing angles, if you know one of the acute angles of a right triangle?

The sum of the measures of all the angles in a triangle is always equal to

180

o

.

In a right triangle, however, one of the angles is already known: the right angle, or the

90

o

angle.

Let the other two angles be

x

and

y

(which will be acute).

Applying these conditions, we can say that,

x

+

y

+

90

o

=

180

o

x

+

y

=

180

o

−

90

o

x

+

y

=

90

o

That is, the sum of the two acute angles in a right triangle is equal to

90

o

.

If we know one of these angles, we can easily substitute that value and find the missing one.

For example, if one of the angles in a right triangle is

25

o

, the other acute angle is given by:

25

o

+

y

=

90

o

y

=

90

o

−

25

o

y

=

65

o

Tanish J. ·  1 · Nov 26 2014

How do you know what trigonometric function to use to solve right triangles?

Right triangles are a special case of triangles. You always know at least one angle, the right angle, and depending on what else you know, you can solve the rest of the triangle with fairly simple formulas.

http://etc.usf.edu/clipart/36500/36521/tri11_36521.htm

If you know any one side and one angle, or any two sides, you can use the pneumonic soh-cah-toa to remember which trig function to use to solve for others.

s

−

i

n

(

θ

)

=

 

o

−

pposite

/

 

h

−

ypotenuse

c

−

o

s

(

θ

)

=

a

−

djacent

/

h

−

ypotenuse

t

−

a

n

(

θ

)

=

o

−

pposite

/

a

−

djacent

Opposite refers to the side which is not part of the angle, adjacent refers to the side that is part of the angle, and the hypotenuse is the side opposite the right angle, which is

C

in the image above.

For example,lets say you know the length of

a

and the value of angle

A

in the above triangle. Using the cosine function you can solve for

c

, the hypotenuse.

cos

(

A

)

=

a

c

Which rearranges to;

c

=

a

cos

(

A

)

If you know the length of both sides

a

and

b

, you can solve for the tangent of either angle

A

or

B

.

tan

(

A

)

=

a

b

Then you take the inverse tangent,

tan

−

1

to find the value of

A

.

Zack M. ·  4 · Dec 7 2014

What are inverse trigonometric functions and when do you use it?

Inverse trigonometric functions are useful in finding angles.

Example

If

cos

θ

=

1

√

2

, then find the angle

θ

.

By taking the inverse cosine of both sides of the equation,

⇒

cos

−

1

(

cos

θ

)

=

cos

−

1

(

1

√

2

)

since cosine and its inverse cancel out each other,

⇒

θ

=

cos

−

1

(

1

√

2

)

=

π

4

I hope that this was helpful.

Wataru ·  1 · Nov 2 2014

What is Solving Right Triangles?

Solving a right triangle means finding missing measures of sides and angles from given measures of sides and angles.

I hope that this was helpful.

Wataru ·  3 · Nov 6 2014

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For a normal distribution, the z value for an x value that is to the right of the mean is always:
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