Find the angle,in degrees,between two adjacent sides of the pyramid shown below.Round to the nearest tenth of a degree.[Note: The base of the pyramid is not considered a side.] P(10,0,0),Q(10,10,0),R(0,10,0),S(5,5,2)
A)2.6°
B)47.1°
C)90°
D)45°
E)59.5°

Question

Find the angle,in degrees,between two adjacent sides of the pyramid shown below.Round to the nearest tenth of a degree.[Note: The base of the pyramid is not considered a side.]  P(10,0,0),Q(10,10,0),R(0,10,0),S(5,5,2)
A)2.6° 
B)47.1° 
C)90° 
D)45° 
E)59.5°

Quizplus · Accepted Answer

To find the angle between two adjacent sides, we need to find the dot product of the two vectors formed by those sides and use the formula $ heta=\cos^{-1}(\frac{\vec{v}\cdot\vec{w}}{|\vec{v}|\cdot|\vec{w}|})$. The two adjacent sides can be PQ and PS, or QR and QS, or RP and RS. We will use PQ and PS as an example. Vector $\vec{v}$ is formed by subtracting the coordinates of point P from those of point Q: $\vec{v}=<0,10,0>$. Vector $\vec{w}$ is formed by subtracting the coordinates of point P from those of point S: $\vec{w}=<-5,5,2>$. To find the dot product of $\vec{v}$ and $\vec{w}$, we multiply the corresponding components and add them up: $\vec{v}\cdot\vec{w}=0*-5+10*5+0*2=50$. To find the magnitudes of $\vec{v}$ and $\vec{w}$, we use the Pythagorean theorem: $|\vec{v}|=\sqrt{0^2+10^2+0^2}=10$ and $|\vec{w}|=\sqrt{(-5)^2+5^2+2^2}=\sqrt{50}$. Plugging in the values into the formula, we get $ heta=\cos^{-1}(\frac{50}{10\sqrt{50}})\approx\boxed{ extbf{(B) } 47.1^\circ}$.

Find the Angle,in Degrees,between Two Adjacent Sides of the Pyramid