Triangle PQR with side p across from angle P, side q across from angle Q, and side r across from angle R If ∠P measures 27°,

4 answers
Question:

Triangle PQR with side p across from angle P, side q across from angle Q, and side r across from angle R
If ∠P measures 27°, q equals 6.0, and r equals 7.2, then which length can be found using the Law of Cosines?
r
q
RQ
PQ

[tex]Triangle PQR with side p across from angle P, side q across from angle Q, and side r across from ang[/tex]


Answers

the answer should be r

Step-by-step explanation:

R

Step-by-step explanation:

Law of cosine expression for the triangle will be

q² = r² + p² - 2 (rp) cos Q

going by the equation r and p is given and Q is included then we can calculate q which corresponds tor R which faces or is subtended by angle Q. p correspond to RQ and r correspond to PQ

= 3.29

Step-by-step explanation:

Let call the length needs to be found is q

We have Law of Cosine:

q = [tex]\sqrt{r^{2} +r^{2} -2prcosP }[/tex]

= [tex]\sqrt{6^{2}+7.2^{2} -2*6*7.2*cos(27)}[/tex]

= 3.29

The length can be found using the Law of Cosines is 3.29

We can find side RQ

Step-by-step explanation:

In order to use the Law of Cosines:

-Two sides and the angle between them must be given to find the third side

-All three sides must be given to find an angle

In this case we are given two sides(q and r) and we are given angle P which is between them. Using this information we can come to the conclusion that side RQ is the length we can find.

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