Let's analyze the two parts of the image and solve for $x$ and $y$ in each triangle problem step by step:

(d) Triangle Problem

In the first triangle (d), we have:

• $\angle P = 70^\circ$,
• $\angle S = 70^\circ$,
• $PQ = 8 \, \text{cm}$,
• $PS = y \, \text{cm}$,
• $PT = 3.2 \, \text{cm}$,
• $ST = 4.3 \, \text{cm}$,
• $QR = 3x - 2 \, \text{cm}$.

This triangle involves the concept of isosceles triangles because $\angle P = \angle S = 70^\circ$. Let’s work through it:

1. Find the third angle $\angle Q$: The sum of the angles in a triangle is $180^\circ$. Therefore: $\angle Q = 180^\circ - 70^\circ - 70^\circ = 40^\circ.$

2. Isosceles triangle properties: Since $\triangle PQS$ is isosceles, $PQ = PS = y$. Hence: $y = 8 \, \text{cm}.$

3. Solve for $x$ using the side ratios: For $\triangle PQR$, use the Law of Sines: $\frac{PQ}{\sin \angle Q} = \frac{QR}{\sin \angle P}.$ Substituting the values: $\frac{8}{\sin 40^\circ} = \frac{3x - 2}{\sin 70^\circ}.$ Solve for $x$:

   • $\sin 40^\circ \approx 0.6428$,
   • $\sin 70^\circ \approx 0.9397$, $\frac{8}{0.6428} = \frac{3x - 2}{0.9397}.$ $12.44 = \frac{3x - 2}{0.9397}.$ Multiply through by $0.9397$: $12.44 \cdot 0.9397 = 3x - 2.$ $11.68 = 3x - 2.$ Add 2 to both sides: $13.68 = 3x.$ Divide by 3: $x = 4.56.$

(e) Triangle Problem

In the second triangle (e), we have:

• $AB = 10 \, \text{cm}$,
• $AC = y + 4.8 \, \text{cm}$,
• $CD = 4.5 \, \text{cm}$,
• $OD = 6 \, \text{cm}$,
• $BO = x \, \text{cm}$.

This involves a right triangle, so we can use the Pythagorean theorem and other geometric properties.

1. Right triangle $\triangle BCD$: In $\triangle BCD$, note that $BD$ is the hypotenuse: $BD^2 = BC^2 + CD^2.$ Substituting $BC = 6 + x$ and $CD = 4.5$: $BD^2 = (6 + x)^2 + 4.5^2.$ $BD^2 = (36 + 12x + x^2) + 20.25.$ $BD^2 = x^2 + 12x + 56.25.$

2. Relate $BD$ to $AC$: Since $BD = AC$, we have: $BD = y + 4.8.$ So: $(y + 4.8)^2 = x^2 + 12x + 56.25.$

3. Simplify further: Expand $(y + 4.8)^2$: $y^2 + 9.6y + 23.04 = x^2 + 12x + 56.25.$

This creates a system of equations. To proceed further, more context might be needed to simplify $x$ and $y$ fully.

Final Answers:

• For (d): $x = 4.56$, $y = 8 \, \text{cm}$.
• For (e): The system of equations requires clarification of constraints to find exact $x$ and $y$.

Would you like detailed clarification for (e)? Or shall we explore related questions?

Related Questions:

1. How do we apply the Law of Sines to non-right triangles?
2. What is the property of angles in an isosceles triangle?
3. How do we use the Pythagorean theorem in right triangles?
4. Can simultaneous equations always give unique solutions?
5. How does geometry relate to trigonometric ratios?

Tip:

Always label triangle sides and angles carefully to avoid confusion when applying formulas.