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 triangle ABC is shown with right angle at B, A at the top right, C at the left, and D on BC between C and B. AC is the slanted hypotenuse labeled 53, BD is the right horizontal segment labeled 15, BC is the full horizontal base, and the shortest distance from D to AC is the perpendicular from D to AC (meeting AC at a right angle) in the center-right area.

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. We need to find: (a) $\angle ACB$ (b) The shortest distance from $D$ to $AC$. --- 2. **Step (a): Find $\angle ACB$** - Since $\angle ABC = 90^\circ$, triangle $ABC$ is right angled at $B$. - Let $BC = x$. Given $BD = \frac{1}{3} BC = 15$ cm, so: $$\frac{1}{3} x = 15 \implies x = 45 \text{ cm}$$ - So, $BC = 45$ cm. - Using Pythagoras theorem in $\triangle ABC$: $$AB^2 + BC^2 = AC^2$$ $$AB^2 + 45^2 = 53^2$$ $$AB^2 + 2025 = 2809$$ $$AB^2 = 2809 - 2025 = 784$$ $$AB = \sqrt{784} = 28 \text{ cm}$$ - Now, to find $\angle ACB$, use trigonometry: $$\tan(\angle ACB) = \frac{AB}{BC} = \frac{28}{45}$$ - Calculate $\angle ACB$: $$\angle ACB = \tan^{-1}\left(\frac{28}{45}\right) \approx 31.9^\circ$$ --- 3. **Step (b): Find the shortest distance from $D$ to $AC$** - Coordinates approach (for clarity): Place $B$ at origin $(0,0)$. Since $BC = 45$ cm along x-axis, $C = (45,0)$. Since $AB = 28$ cm along y-axis, $A = (0,28)$. - Point $D$ lies on $BC$ such that $BD = 15$ cm, so $D = (15,0)$. - Equation of line $AC$: Slope $m = \frac{28 - 0}{0 - 45} = \frac{28}{-45} = -\frac{28}{45}$ Equation: $$y - 28 = -\frac{28}{45}(x - 0) \implies y = -\frac{28}{45}x + 28$$ - Distance from point $D(15,0)$ to line $AC$: Formula for distance from point $(x_0,y_0)$ to line $Ax + By + C = 0$ is: $$d = \frac{|Ax_0 + By_0 + C|}{\sqrt{A^2 + B^2}}$$ - Rewrite line $AC$ in standard form: $$y = -\frac{28}{45}x + 28 \implies \frac{28}{45}x + y - 28 = 0$$ Multiply both sides by 45 to clear fraction: $$28x + 45y - 1260 = 0$$ - Calculate distance: $$d = \frac{|28(15) + 45(0) - 1260|}{\sqrt{28^2 + 45^2}} = \frac{|420 - 1260|}{\sqrt{784 + 2025}} = \frac{840}{\sqrt{2809}}$$ $$\sqrt{2809} = 53$$ $$d = \frac{840}{53} \approx 15.8 \text{ cm}$$ --- **Final answers:** (a) $\angle ACB \approx 31.9^\circ$ (b) Shortest distance from $D$ to $AC$ is approximately $15.8$ cm.