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TRIG HELP!!

Find the value of sin P rounded to the nearest hundredth, if necessary.

TRIG HELP!! Find the value of sin P rounded to the nearest hundredth, if necessary-example-1
User Aledujke
by
7.6k points

2 Answers

2 votes

Answer:

0.38 (rounded)

Explanation:

We can use the Pythagorean theorem to find the hypotenuse.


5^2+12^2 = c^2\\25 +144 = c^2\\169 = c^2\\13 = c

We can now find sin(p). Recall sin is the "opposite over hypotenuse."


sin((5)/(13)) = 0.375

User Carlos Grossi
by
8.1k points
0 votes

Answer:


\sf sin\: P = (5 )/( 13)

Explanation:

Given:

In ∆ PQR

  • PQ = 12
  • QR = 5

To find:

  • sin P

Solution:

To find sin P, we can use the following formula:


\boxed{\boxed{\sf sin \: P =(Opposite )/(Hypotenuse) = (QR )/( PR)}}

We know that QR = 5 and that PR = the hypotenuse of the triangle. We can use the Pythagorean Theorem to find PR:


\begin{aligned} PR^2 & = PQ^2 + QR^2 \\\\ PR^2 & = 12^2 + 5^2 \\\\ PR^2 & = 169 \\\\ PR & =√(169)\\\\ PR & = 13 \end{aligned}

Now that we know the values of QR and PR, we can plug them into the formula for sin P:


\sf sin\: P =( QR )/(PR) = (5 )/( 13)

Therefore,
\sf sin\: P = (5 )/( 13)

User Adalle
by
7.5k points

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