. 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
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 $\boxed{\textbf{(C) }4\sqrt{6}-4\sqrt{2}}$.
To find the length of |RP| in triangle PQR, we can use the Law of Cosines. The Law of Cosines states that in a triangle, the square of one side is equal to the sum of the squares of the other two sides, minus twice their product, multiplied by the cosine of the angle between them.
Let's denote |RP| as c, and |RQ| as a. Therefore, |PQ| is the remaining side and will be denoted as b.
Given that |RQ| = 4 and angle RPQ = 45 degrees, we can plug these values into the Law of Cosines to solve for |RP|.
c^2 = a^2 + b^2 - 2ab * cos(RPQ)
|RP|^2 = 4^2 + b^2 - 2 * 4 * b * cos(45)
|RP|^2 = 16 + b^2 - 8b * cos(45)
Since angle PQR = 60 degrees, we know that angle RPQ = angle RQP = 180 - 45 - 60 = 75 degrees. Therefore, cos(75) = -cos(45).
|RP|^2 = 16 + b^2 + 8b * cos(45)
We also have the relationship that the sum of the angles in a triangle is 180 degrees. Therefore, angle PQR = 60 degrees and angle PRQ = 180 - 45 - 60 = 75 degrees.
Now, we can use the Law of Sines to relate the lengths of the sides to the angles:
sin(PQR) / PQ = sin(PRQ) / PR
sin(60) / b = sin(75) / |RP|
sin(60) / b = -sin(45) / |RP|
Now we can solve for b in terms of |RP|:
b = |RP| * (sin(60) / -sin(45))
b = -|RP| * (sqrt(3) / sqrt(2))
Now, let's substitute the value of b into the equation for |RP|^2:
|RP|^2 = 16 + (-|RP| * (sqrt(3) / sqrt(2)))^2 + 8 * (-|RP| * (sqrt(3) / sqrt(2))) * cos(45)
Simplifying:
|RP|^2 = 16 + |RP|^2 * (3/2) + 8 * |RP| * (sqrt(3) / sqrt(2)) * cos(45)
Multiplying by 2 to eliminate the fraction:
2 * |RP|^2 = 32 + 3 * |RP|^2 + 16 * |RP| * (sqrt(3) / sqrt(2)) * cos(45)
2 * |RP|^2 - 3 * |RP|^2 = 32 + 16 * |RP| * (sqrt(3) / sqrt(2)) * cos(45)
-|RP|^2 = 32 + 16 * |RP| * (sqrt(3) / sqrt(2)) * cos(45)
Now, let's substitute cos(45) = sqrt(2)/2:
-|RP|^2 = 32 + 16 * |RP| * (sqrt(3) / sqrt(2)) * (sqrt(2)/2)
-|RP|^2 = 32 + 16 * |RP| * (sqrt(3))
Now, we can solve for |RP|:
16 * |RP| * (sqrt(3)) = -|RP|^2 - 32
Dividing both sides by |RP|:
16 * sqrt(3) = -|RP| - 32 / |RP|
Multiplying both sides by |RP|:
16 * sqrt(3) * |RP| = -|RP|^2 - 32
Rearranging the terms:
|RP|^2 + 16 * sqrt(3) * |RP| + 32 = 0
We can now solve this quadratic equation using the quadratic formula:
|RP| = (-b ± sqrt(b^2 - 4ac)) / 2a
In this case, a = 1, b = 16 * sqrt(3), and c = 32.
|RP| = (-16 * sqrt(3) ± sqrt((16 * sqrt(3))^2 - 4 * 1 * 32)) / 2 * 1
|RP| = (-16 * sqrt(3) ± sqrt(768 - 128)) / 2
|RP| = (-16 * sqrt(3) ± sqrt(640)) / 2
|RP| = (-16 * sqrt(3) ± 8 * sqrt(10)) / 2
|RP| = -8 * sqrt(3) ± 4 * sqrt(10)
Since |RP| cannot be negative, we can discard the negative solution.
Therefore, we have |RP| = 4 * sqrt(10).
So, the correct answer is D) 4 * sqrt(10).