The diagram is that of a light inextensible string of length 4.2m, whose ends are attached to two fixed points X and Y, 3m apart, and on the same horizontal...

The geometry of the string gives \(XO=4.2-2.4=1.8\,\text{m}\), \(OY=2.4\,\text{m}\), and \(XY=3.0\,\text{m}\).

(a) Angle \(XOY\)

\[ XO^2+OY^2=1.8^2+2.4^2=3.24+5.76=9=3^2=XY^2. \]

Therefore, by the converse of Pythagoras’ theorem,

\[ \angle XOY=90^\circ. \]

(b) Horizontal force \(P\), and (c) tension \(T\)

The two tensions have equal magnitude \(T\). From the right-angled triangle:

• For \(OX\), the horizontal and vertical components of tension are \(0.6T\) to the left and \(0.8T\) upwards.
• For \(OY\), the horizontal and vertical components of tension are \(0.8T\) to the right and \(0.6T\) upwards.

The mass is \(0.800\,\text{kg}\), so its weight is \(W=0.8g\).

Resolving vertically for equilibrium:

\[ 0.8T+0.6T=0.8g \]

\[ 1.4T=0.8g \]

\[ T=\frac{0.8g}{1.4}=\frac{4g}{7}\,\text{N}. \]

Taking \(g=10\,\text{m s}^{-2}\),

\[ T=\frac{40}{7}=5.71\,\text{N}. \]

Horizontally, the rightward component is \(0.8T\) and the leftward component is \(0.6T\). Their resultant is \(0.2T\) to the right, so \(P\) must act to the left:

\[ P=0.8T-0.6T=0.2T. \]

\[ P=0.2\left(\frac{40}{7}\right)=\frac{8}{7}=1.14\,\text{N}. \]

Thus:

• \(\angle XOY=90^\circ\)
• \(P=1.14\,\text{N}\), acting horizontally towards \(X\)
• \(T=5.71\,\text{N}\)

The supplied school reference answer is not consistent with equilibrium: it treats \(P\) as though it were found directly from a string length, and then combines \(P\) and the weight to obtain one tension. There are two equal tension forces, so their horizontal and vertical components must both be included when resolving forces.

Examination reminder: Draw the force directions at the hanging body, resolve both tensions into horizontal and vertical components, and apply equilibrium separately in each direction.