We can use the Law of Cosines to find |RP|:
$|RP|^2 = |PQ|^2 + |QR|^2 - 2|PQ||QR|\cos\angle PQR$
We know that |QR| = 4 and $\angle PQR = 60^\circ$, so $\cos\angle PQR = \frac{1}{2}$.
To find |PQ|, we can use the Law of Sines:
$\frac{|PQ|}{\sin\angle RPQ} = \frac{|QR|}{\sin\angle QRP}$
$\frac{|PQ|}{\sin 45^\circ} = \frac{4}{\sin 75^\circ}$
$|PQ| = \frac{4\sin 45^\circ}{\sin 75^\circ}$
$|PQ| = \frac{4}{\sqrt{3}+\sqrt{2}} = 4(\sqrt{3}-\sqrt{2})$
Now we can substitute |PQ| and |QR| into the Law of Cosines and simplify:
$|RP|^2 = (4(\sqrt{3}-\sqrt{2}))^2 + 4^2 - 2(4(\sqrt{3}-\sqrt{2}))(4)(\frac{1}{2})$
$|RP|^2 = 32 - 16\sqrt{6} + 16 + 16$
$|RP|^2 = 64 - 16\sqrt{6}$
$|RP| = \sqrt{64 - 16\sqrt{6}} = \sqrt{16(4-\sqrt{6})}$
$|RP| = 4\sqrt{4-\sqrt{6}}$
Therefore, the answer is $\boxed{\textbf{(C) }4\sqrt{6}-4\sqrt{2}}$.
. In the triangle PQR below, angle RPQ=45,
angle PQR=60 and |RQ|=4. Find |RP|
A. √26
B. 3√6
C. 4√
6
3
D. 2√6
1 answer