Two ships P and Q left port at the same time. Q sailed at a bearing of 150 degrees while P sailed on the north side of Q. After a distance of 8km and 10km by P and Q respectively, their distance apart is 12km. Find the bearing of P from R

If the identities or equations are applicable for all the triangles and not just for right triangles, then they are the triangle identities. These identities will include:

Sine law

Cosine law

Tangent law

According to question,

The angle between RP & RQ.

Using the cosine law

[tex]PQ^{2} = RP^{2} + QR^{2} - 2PR*RQ*cosx\\12^{2}=8^2+10^2-2*8*10cosx\\cosx = \frac{164-144}{160} \\cosx = \frac{20}{160} \\cosx =\frac{1}{8}[/tex]

[tex]x=83[/tex] degree

[tex]150-83=67[/tex]

Hence, the bearing of P from R is 67 degree

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Two ships P and Q left port at the same time. Q sailed at a bearing of 150 degrees while P sailed on the north side of Q. After a distance of 8km and 10km by P and Q respectively, their distance apart is 12km. Find the bearing of P from R​

Answer :

What is trigonometry identities.

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Two ships P and Q left port at the same time. Q sailed at a bearing of 150 degrees while P sailed on the north side of Q. After a distance of 8km and 10km by P and Q respectively, their distance apart is 12km. Find the bearing of P from R