Question: 4 2 In triangle ABC, angle ABC = 90°. D is a point on BC such that BD is 1/3 of BC. BD is 15 cm and AC is 53 cm. (a) Find angle ACB. Answer ∠ACB = 31.9° (b) Calculate the shortest length from D to AC. Answer 15.8 cm Graph/shape: A right-angled triangle ABC with the right angle at B, A at the upper right, C at the left, and D on the base BC between C and B; segment AD is drawn inside the triangle, BC is horizontal, AB is vertical, and AC is the slanted hypotenuse. position_hint: center

1. **Problem statement:** We have a right triangle $ABC$ with $\angle ABC = 90^\circ$. Point $D$ lies on $BC$ such that $BD$ is $\frac{1}{3}$ of $BC$. Given $BD = 15$ cm and hypotenuse $AC = 53$ cm. 2. **Find $\angle ACB$ (angle at $C$).** - Since $BD = \frac{1}{3} BC$ and $BD = 15$, then $BC = 3 \times 15 = 45$ cm. - Triangle $ABC$ is right angled at $B$, so by Pythagoras theorem: $$AB = \sqrt{AC^2 - BC^2} = \sqrt{53^2 - 45^2} = \sqrt{2809 - 2025} = \sqrt{784} = 28 \text{ cm}.$$ - To find $\angle ACB$, use the sine or cosine rule. Using sine: $$\sin \angle ACB = \frac{AB}{AC} = \frac{28}{53}.$$ - Calculate the angle: $$\angle ACB = \arcsin\left(\frac{28}{53}\right) \approx 31.9^\circ.$$ 3. **Calculate the shortest length from $D$ to $AC$.** - Coordinates approach: Place $B$ at origin $(0,0)$, $C$ at $(45,0)$, and $A$ at $(0,28)$. - Point $D$ lies on $BC$ at $BD=15$, so $D$ is at $(15,0)$. - Line $AC$ passes through $A(0,28)$ and $C(45,0)$. - Equation of line $AC$: $$\text{slope} = m = \frac{0-28}{45-0} = -\frac{28}{45}.$$ $$y - 28 = -\frac{28}{45}(x - 0) \Rightarrow y = 28 - \frac{28}{45}x.$$ - Distance from point $D(15,0)$ to line $AC$: $$\text{Distance} = \frac{|m x_0 - y_0 + b|}{\sqrt{m^2 + 1}}$$ where line in form $y = mx + b$, here $b=28$. Substitute: $$\text{Distance} = \frac{|-\frac{28}{45} \times 15 - 0 + 28|}{\sqrt{\left(-\frac{28}{45}\right)^2 + 1}} = \frac{| -\frac{420}{45} + 28|}{\sqrt{\frac{784}{2025} + 1}} = \frac{| -9.333 + 28|}{\sqrt{1.386}} = \frac{18.667}{1.177} \approx 15.8 \text{ cm}.$$