Ask question

Ask question

All categories

3 years ago

14

For triangle ABC, find the measure of AB given me∠ A = 55 degrees, m∠b= 44 degrees, and b=6. . A. 45.22. B.96.68. C.88.19. .

2 answers:

jekas [21]3 years ago

4 0

There exist a similar question where b = 68 instead of 6. First, determine the measure of the third angle, angle C,
m∠c = 180 - (55° + 44°) = 81°
Let x be the side AB, that which is opposite to angle C. Through the Sine Law,
68 / sin 44° = x / sin 81°
From the equation, the value of x is equal to 96.68. Thus, the answer is letter B.

Sever21 [200]3 years ago

3 0

Answer:

Here are some answers to the waca honors geometry law of sines and law of cosines quiz:

1. B

2. B

3. B

4. D

5. D

6. B

7. C

Sorry, I know this ain't helpful to any people doing non honors

Step-by-step explanation:

You might be interested in

The expression 2y+3y2−5y+y2simplifies to 4y2−3y <br> true or false

Mumz [18]

Answer:

False?

Step-by-step explanation:

I'm pretty sure it's false because wouldn't it simplify to 5y? Sorry if I'm wrong. :)

8 0

3 years ago

How do i graph 2y=1/2x+10 using slope intercept form

sergij07 [2.7K]

2y=1/2x+10
divide by two
y=1/4x+5

3 0

3 years ago

In the figure, PQ is parallel to RS. The length of RP is 2 cm; the length of PT is 18 cm; the length of QT is 27 cm. What is the

Sidana [21]

Step 1

<u>Find the value of TS</u>

we know that

if PQ is parallel to RS. then triangles TRS and TPQ are similar

so

$\frac{TR}{TP} =\frac{TS}{QT}$

solve for TS

$TS =\frac{TR*QT}{TP}$

we have

$RP=2\ cm\\TP=18\ cm\\QT=27\ cm$

$TR=TP+RP\\TR=18+2=20\ cm$

substitute

$TS =\frac{20*27}{18}$

$TS =30\ cm$

Step 2

<u>Find the value of SQ</u>

we know that

$SQ=TS-QT$

we have

$TS =30\ cm$

$QT=27\ cm$

substitute

$SQ=30\ cm-27\ cm=3\ cm$

therefore

<u>the answer is</u>

the value of SQ is $3\ cm$

6 0

3 years ago

Read 2 more answers

ASAP please help me I need this quick <br> The question and answers are in the picture

Ilia_Sergeevich [38]

Answer:

25

Step-by-step explanation:

maybe this will help you

4 0

3 years ago

Read 2 more answers

Theorem: A line parallel to one side of a triangle divides the other two proportionately.

Kisachek [45]

Answer:

Segment BF = 16

Step-by-step explanation:

The given theorem states that a line parallel to one side of a triangle divides the other two sides proportionately

The given theorem is the Triangle Proportionality Theorem

According to the theorem, given that segment DE is parallel to segment BC, we have;

$\dfrac{AD}{BD} = \dfrac{AE}{EC}$

Therefore;

$BD = \dfrac{AD}{\left(\dfrac{AE}{EC} \right) } = AD \times \dfrac{EC}{AE}$

Which gives;

$BD = 6 \times \dfrac{18}{12}= 9$

Similarly, given that EF is parallel to AB, we get;

$\dfrac{AE}{EC} = \dfrac{BF}{FC}$

Therefore;

$BF = FC \times \dfrac{AE}{EC}$

Which gives;

$BF = 24 \times \dfrac{12}{18} = 16$

Therefore, the statement that can be proved using the given theorem is segment BF = 16.

8 0

3 years ago