Here are the calculations for the unknown values in each diagram: a) Step 1: Identify the type of triangle. The tick marks on sides NP and PO indicate that $NP = PO$. Therefore, $\triangle NPO$ is an isosceles triangle. Step 2: Apply properties of isosceles triangles. In an isosceles triangle, the angles opposite the equal sides are equal. $$ \angle PNO = \angle PON = 46^\circ $$ Thus, $x = 46^\circ$. Step 3: Apply the sum of angles in a triangle. The sum of angles in $\triangle NPO$ is $180^\circ$. $$ \angle NPO + \angle PNO + \angle PON = 180^\circ $$ $$ y + 46^\circ + 46^\circ = 180^\circ $$ $$ y + 92^\circ = 180^\circ $$ $$ y = 180^\circ - 92^\circ $$ $$ y = 88^\circ $$ The unknown values are: $$ \boxed{x = 46^\circ, y = 88^\circ} $$ b) Step 1: Identify the given angles. In $\triangle RST$, $\angle R = 34^\circ$ and $\angle S = 72^\circ$. Step 2: Apply the exterior angle theorem. The exterior angle $x$ at vertex T is equal to the sum of the two opposite interior angles. $$ x = \angle R + \angle S $$ $$ x = 34^\circ + 72^\circ $$ $$ x = 106^\circ $$ The unknown value is: $$ \boxed{x = 106^\circ} $$ c) Step 1: Find angle $y$. The angle $y$ and the $67^\circ$ angle form a linear pair on a straight line. Angles on a straight line sum to $180^\circ$. $$ y + 67^\circ = 180^\circ $$ $$ y = 180^\circ - 67^\circ $$ $$ y = 113^\circ $$ Step 2: Find angle $x$. The sum of angles in $\triangle UTW$ is $180^\circ$. $$ \angle U + \angle UTW + \angle TWU = 180^\circ $$ $$ 59^\circ + x + y = 180^\circ $$ Substitute the value of $y$: $$ 59^\circ + x + 113^\circ = 180^\circ $$ $$ x + 172^\circ = 180^\circ $$ $$ x = 180^\circ - 172^\circ $$ $$ x = 8^\circ $$ The unknown values are: $$ \boxed{x = 8^\circ, y = 113^\circ} $$ d) Step 1: Analyze the given triangles. We have two right-angled triangles: $\triangle HLN$ (right-angled at L) and $\triangle SRT$ (right-angled at R). From the angle markings: $\angle H = \angle S$ (one arc) $\angle N = \angle T$ (two arcs) $\angle L = \angle R = 90^\circ$ (right angle symbol) Step 2: Determine similarity. Since all three corresponding angles are equal, the two triangles are similar by the AAA (Angle-Angle-Angle) similarity criterion. The correspondence of vertices is $H \leftrightarrow S$, $L \leftrightarrow R$, $N \leftrightarrow T$. Step 3: Set up ratios of corresponding sides. For similar triangles, the ratio of corresponding sides is equal: $$ \frac{HL}{SR} = \frac{HN}{ST} $$ Step 4: Substitute known values and solve for $x$. Given: $HL = 15$ cm, $HN = 18$ cm, $ST = 24$ cm, $SR = x$ cm. $$ \frac{15}{x} = \frac{18}{24} $$ Simplify the ratio on the right side: $$ \frac{18}{24} = \frac{3 \times 6}{4 \times 6} = \frac{3}{4} $$ So, the equation becomes: $$ \frac{15}{x} = \frac{3}{4} $$ Cross-multiply: $$ 15 \times 4 = 3 \times x $$ $$ 60 = 3x $$ $$ x = \frac{60}{3} $$ $$ x = 20 $$ The length is in cm. The unknown value is: $$ \boxed{x = 20 \text{ cm}} $$ That's 2 down. 3 left today — send the next one.