Question(9) If : $\cos \frac{20^{\circ}+\theta}{2}=\sin \frac{40^{\circ}+\theta}{2}, 0^{\circ}<\theta<90^{\circ}$ then $\theta=$ $\qquad$ (a) $60^{\circ}$ (b) $45^{\circ}$ (c) $30^{\circ}$ (d) $20^{\circ}$ (10) If the ratio between areas of two similar triangles equals $9: 25$ and the pe the smaller triangle is 60 cm then the perimeter of the greater triangle equals (a) 60 (b) 80 (c) 100 (d) 120

Studdy Solution
For problem (10), use the relationship between the areas and perimeters of similar triangles.
Given the ratio of areas is $9:25$, the ratio of corresponding sides is the square root of the area ratio: $$\sqrt{\frac{9}{25}} = \frac{3}{5}$$
Let $P_s$ be the perimeter of the smaller triangle and $P_g$ be the perimeter of the greater triangle. The ratio of perimeters is the same as the ratio of corresponding sides: $$\frac{P_s}{P_g} = \frac{3}{5}$$
Given $P_s = 60$ cm, solve for $P_g$: $$\frac{60}{P_g} = \frac{3}{5}$$
Cross-multiply to solve for $P_g$: $$3P_g = 5 \times 60$$
$$3P_g = 300$$
Divide by 3: $$P_g = 100$$
The solutions are: For problem (9), $\theta = 60^\circ$. For problem (10), the perimeter of the greater triangle is $100$ cm.