Use resultant formula
[tex]\boxed{\sf R=\sqrt{A^2+B^2+2ABcos\theta}}[/tex]
So
#1
A be p+q and B be p-q
[tex]\\ \rm\Rrightarrow R=\sqrt{3p^2+q^2}[/tex]
[tex]\\ \rm\Rrightarrow \sqrt{(p+q)^2+(p-q)^2+2(p+q)(p-q)cos\alpha}=\sqrt{3p^2+q^2}[/tex]
[tex]\\ \rm\Rrightarrow 2p^2+2q^2+2(p^2-q^2)cos\alpha=3p^2+q^2[/tex]
[tex]\\ \rm\Rrightarrow 2cos\alpha=1[/tex]
[tex]\\ \rm\Rrightarrow cos\alpha=\dfrac{1}{2}[/tex]
[tex]\\ \rm\Rrightarrow \alpha=\dfrac{\pi}{3}[/tex]
#2
[tex]\\ \rm\Rrightarrow 2(p^2+q^2)+2(p^2-q^2)cos\beta=2(p^2+q^2)[/tex]
[tex]\\ \rm\Rrightarrow 2cos\beta=0[/tex]
[tex]\\ \rm\Rrightarrow cos\beta=0[/tex]
[tex]\\ \rm\Rrightarrow \beta=\dfrac{\pi}{2}[/tex]
#3
[tex]\\ \rm\Rrightarrow 2(p^2+q^2)+2(p^2-q^2)cos\gamma=p^2+q^2[/tex]
[tex]\\ \rm\Rrightarrow 2(p^2-q^2)cos\gamma=-(p^2+q^2)[/tex]
[tex]\\ \rm\Rrightarrow cos\gamma =\dfrac{q^2-p^2}{2(p^2-q^2)}[/tex]
[tex]\\ \rm\Rrightarrow \gamma=cos^{-1}\left(\dfrac{q^2-p^2}{2(p^2+q^2)}\right)[/tex]
Answer:
See below ~
Explanation:
Here, let's apply the resultant vector formula :
[tex]R = \sqrt{A^{2} + B^{2} + 2ABcos\theta}[/tex]
Part (i) :
√3P² + Q² = √(P + Q)² + (P - Q)² + 2(P + Q)(P - Q)cosθ
3P² + Q² = P² + Q² + 2PQ + P² + Q² - 2PQ + 2(P² - Q²)cosθ
2(P² - Q²)cosθ = P² - Q²
cosθ = 1/2
θ = π/3 or 60°
Part (ii) :
√2(P² + Q²) = √(P + Q)² + (P - Q)² + 2(P + Q)(P - Q)cosθ
2(P² + Q²) = P² + Q² + 2PQ + P² + Q² - 2PQ + 2(P² - Q²)cosθ
2(P² - Q²)cosθ = 0
cosθ = 0
θ = π/2 or 90°
Part (iii) :
√P² + Q² = √(P + Q)² + (P - Q)² + 2(P + Q)(P - Q)cosθ
P² + Q² = P² + Q² + 2PQ + P² + Q² - 2PQ + 2(P² - Q²)cosθ
2(P² - Q²)cosθ = -P² - Q²
cosθ = [tex]\frac{-(P^{2}+ Q^{2}) }{2(P^{2}-Q^{2}) }[/tex]
[tex]\theta = cos^{-1} \frac{-(P^{2}+ Q^{2}) }{2(P^{2}-Q^{2}) }[/tex]
