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masya89 [10]
2 years ago
10

Can someone help me with this question

Mathematics
1 answer:
gavmur [86]2 years ago
3 0

Answer:

w = 2.6

x = 26.2^\circ

Step-by-step explanation:

Solving for w:

Look at the right triangle on the left side. It has a 32-deg angle, a hypotenuse of length 5, and a leg of length w. For the 32-deg angle, w is the opposite leg. We have an opposite leg and a hypotenuse, so we use the sine.

\sin A = \dfrac{opp}{hyp}

\sin 32^\circ = \dfrac{w}{5}

w = 5 \sin 32^\circ

w = 2.6495

w = 2.6

Solving for x:

Now look at the right triangle on the right side. There is an acute angle measuring x, an opposite leg measuring 2.6495 (from the solution above), and a hypotenuse measuring 6. We use the sine again.

\sin A = \dfrac{opp}{hyp}

\sin x = \dfrac{w}{6}

\sin x = \dfrac{2.6495}{6}

\sin x = 0.441599

Since we know what the sine of angle x is equal to, and we want to know the measure of angle x, we use the inverse sine function, also called the arcsine.

x = \sin^{-1} 0.441599

x = 26.2^\circ

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Answer:

c) 12/15 = 4/5

Step-by-step explanation:

imagine we mirror the triangle up, so that Z is on top.

then you can clearly see that 6 is cos(X) times r (and r is then 7.5).

XY is sin(X)×7.5

and again, 7.5 is r (the line making the X angle).

so, the cosine ratio of X is

6 = cos(X)×7.5

cos(X) = 6/7.5 or then 12/15. or simplified 4/5.

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2 years ago
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If QS bisects PQT, SQT=(8x- 25),PQT=(9x+34), and SQR=112, find each measure.
Goryan [66]

The values of the angles are; x = 12°, m∠PQS = 71°, m∠PQT = 142° and m∠TQR = 41°

<h3>What are the measure of the each angle?</h3>

The required angles; x, m∠PQS, m∠PQT, and m∠TQR

The given parameters are bisects ∠PQT

m∠SQT = (8·x - 25)°

m∠PQT = (9·x + 34)°

m∠SQR = 112°

We have;

m∠PQT = m∠SQT + m∠PQS (Angle addition postulate)

m∠SQT ≅ m∠PQS (Angles formed by angle bisector are congruent)

m∠SQT = m∠PQS  by Definition of congruency

m∠PQT = 2 × m∠SQT

Therefore;

(9·x + 34)° = 2 × (8·x - 25)° = (16·x - 50)°

Collecting like terms gives:

(34 + 50)° = 16·x - 9·x = 7·x

7·x = 84°

x = 84°/7 = 12°

x = 12°

m∠SQT = (8·x - 25)°

Therefore;

m∠SQT = (8 × 12 - 25)° = 71°

m∠SQT = 71°

m∠PQT = 2 × m∠SQT

∴ m∠PQT = 2 × 71° = 142°

m∠PQT = 142°

m∠PQS = m∠SQT (Angles formed by the same bisector )

∴ m∠PQS = m∠SQT = 71°

m∠PQS = 71°

m∠SQR = m∠SQT + m∠TQR (Angle addition postulate)

m∠SQT = 71°

∴  m∠SQR = 112° = 71° + m∠TQR

m∠TQR = 112° - 71° = 41°

m∠TQR = 41°

Learn more about angles here:

brainly.com/question/2882938

#SPJ1

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