In PQR, Q= 76, PQ = 18, and QR = 27
Please find angle P, T and PR for 5

In PQR, Q= 76, PQ = 18, And QR = 27 Please Find Angle P, T And PR For 5

Answers

Answer 1

Answer:

Step-by-step explanation:

Given:

m<Q = 76°

PQ = r = 18

QR = p = 27

Required:

Find P, T, and PR (q)

Solution:

✔️To find PR(q), apply the Law of Cosines.

Thus:

q² = r² + p² - 2rp×cos(Q)

Plug in the values

q² = 18² + 27² - 2×18×27×cos(76)

q² = 1,053 - 235.148

q² = 817.852

q = √817.852

q = 28.6 (nearest tenth)

✔️Find P by applying the Law of Sines:

[tex] \frac{sin(P)}{p} = \frac{sin(Q)}{q} [/tex]

Plug in the values

[tex] \frac{sin(P)}{27} = \frac{sin(76)}{28.6} [/tex]

[tex] sin(P) = \frac{sin(76)*27}{28.6} [/tex]

[tex] sin(P) = 0.9160 [/tex]

[tex] P = sin^{-1}(0.9160) [/tex]

P = 66.3° (nearest tenth)

✔️R = 180 - (P + Q)

R = 180 - (66.3 + 76)

R = 37.7°


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nPr = n! / (n-r)!

7!
------
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=
7!
----
5!

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Step-by-step explanation:

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Answers

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I'm on edenuity too

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Step-by-step explanation:

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Answers

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Answers

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