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Solve by using the measurements ∠PQR = 90°, ∠QRP = 80°, and r = 15. Round measures of sides to the nearest tenth and measures of angles to the nearest degree.

Solve by using the measurements ∠PQR = 90°, ∠QRP = 80°, and r = 15. Round measures-example-1
User Swordstoo
by
6.4k points

2 Answers

5 votes

Answer:

q ≈ 15.2

p ≈ 2.5

Explanation:

Given that:

  • ∠PQR = 90°
  • ∠QRP = 80°
  • r = 15

We need to find the length of side q and p

Use the sine law to find out q, we have:

sin(QRP) =
(r)/(q)

<=> sin(80°) =
(r)/(q)

<=> q =
(r)/(sin(80)) =
(15)/(0.98) ≈ 15.2

Because PQR is a right triangle, so we use pytagon theorem to find p


q^(2) = r^(2) + p^(2)

<=>
p^(2) = q^(2) - r^(2)

<=>
p^(2) = 15.2^(2) - 15^(2) = 6.04 \\

<=> p =
√(6.04) ≈ 2.5

User Polyana Fontes
by
5.7k points
7 votes

Good evening ,

___________________

Explanation:

Look at the photo below for the answer.

:)

Solve by using the measurements ∠PQR = 90°, ∠QRP = 80°, and r = 15. Round measures-example-1
User Lucca Mordente
by
5.5k points