Geometry Question

Question

anyone can help me with this question: in the triangle, ABC below points A and C are fixed however point B moves in time by a constant increment h, if you know α angle and segment AB, calculate the segments AB', AB''.

Answer 1

1) Find $\angle ABC$ in terms of $\alpha$.

Added:- By angle sum of triangle, $\beta = 90^ - 0.5\alpha$.

2) Find AC by sine law. Then, BC is known.

Added:- Let AB be k units, a known quantity. $AC = \dfrac {k \sin \beta}{\sin \alpha}$. Then, $BC = AC$.

3) AB’ can be found by applying cosine law.

Added:- In $\triangle ACB’$, $AB’ = \sqrt {(h + AC)^2 + (AC)^2 – 2(h + AC)(AC) \cos \alpha}$.

Answer 2

(That's my first answer here)

Usually the angle $\alpha$ is next to the point $A$ and the angle $\gamma$ is next to the point $C$.
So I'll call the given angle $\gamma$ instead of $\alpha$.
Further more I'll define the points $$ B_n := \begin{cases} B'\ , & \text{if $n$ is 1} \\ B'', & \text{if $n$ is 2.} \end{cases} $$.

I'll give you a short OVERVIEW about how to calculate the desired segments $\overline{A B'}$ and $\overline{A B''}$:

1. Calculate the altitude $h_a$ to the side $\overline{BC}$.
   Let's call the intersection of $h_a$ and $\overline{BC}$: $$H_a := h_a \cap \overline{BC}$$
2. Calcualte the length of $\overline{H_a B}$.
3. Calcualte the length of $\overline{H_a B_n}$.
4. Calcualte the length of $\overline{A B_n}$.

DETAILS:

1. Calculate $h_a$ by applying the 'Law of sines' to the triangle $\triangle AH_aC$: $$ \begin{align} \frac{h_a}{\sin \gamma} & = \frac{\overline{AC}}{\sin \frac \pi 2} \tag{$\sin \frac \pi 2 = 1$} \\ \frac{h_a}{\sin \gamma} & = \overline{AC} \tag{multiply with $\sin \gamma$} \\ h_a & = \overline{AC} \cdot \sin \gamma \end{align} $$

2. Calcualte the length of $\overline{H_a B}$ by applying the Pythagorean theorem on the triangle $\triangle H_a A B$: $$ \begin{align} \overline{H_a B}^2 & = \overline{AB}^2 - h_a^2 \tag{2a} \\ \overline{H_a B} & = \sqrt{\overline{AB}^2 - h_a^2} \tag{2b} \end{align} $$

3. Calcualte the length of $\overline{H_a B_n}$: $$ \begin{align} \overline{H_a B_n} & = \overline{H_a B} + \overline{B B_n} \\ & = \overline{H_a B} + n \cdot h \end{align} $$

4. Calcualte the length of $\overline{A B_n}$ by applying the Pythagorean theorem on the triangle $\triangle B_n H_a A$: $$ \begin{align} \overline{A B_n} & = \sqrt{h_a^2 + \overline{H_a B_n}^2} \tag{apply step 3} \\ & = \sqrt{h_a^2 + \left(\overline{H_a B} + n \cdot h\right)^2} \\ & = \sqrt{h_a^2 + \overline{H_a B}^2 + 2 n h \overline{H_a B} + n^2 h^2} \tag{apply steps 2a and 2b} \\ & = \sqrt{h_a^2 + \overline{AB}^2 - h_a^2 + 2 n h \sqrt{\overline{AB}^2 - h_a^2} + n^2 h^2} \tag{simplify} \\ & = \sqrt{\overline{AB}^2 + 2 n h \sqrt{\overline{AB}^2 - h_a^2} + n^2 h^2} \tag{apply step 1} \\ & = \sqrt{\overline{AB}^2 + 2 n h \sqrt{\overline{AB}^2 - \overline{AC}^2 (\sin \gamma) ^2} + n^2 h^2} \\ \end{align} $$

   4a. Calculate Length of $\overline{A B'}$ by setting $n = 1$: $$ \overline{A B'} = \sqrt{\overline{AB}^2 + 2 h \sqrt{\overline{AB}^2 - \overline{AC}^2 (\sin \gamma) ^2} + h^2} $$ 4b. Calculate Length of $\overline{A B''}$ by setting $n = 2$: $$ \overline{A B'} = \sqrt{\overline{AB}^2 + 4 h \sqrt{\overline{AB}^2 - \overline{AC}^2 (\sin \gamma) ^2} + 4h^2} $$

Answer 3

Write $\overline{AB}=r$, $\overline{AC}=\overline{BC}=s$, and $\overline{BB'}=x$. Then by the law of cosines, $$r^2 = s^2 + s^2 - 2s^2\cos\alpha\quad\Rightarrow\quad s = \frac{r}{\sqrt{2-2\cos\alpha}}.$$ Next, consider $\triangle AB'C$. There, $$\overline{AB'} = s^2 + (s+x)^2 - 2s(s+x)\cos\alpha.$$ Substitute the value of $s$ above into this equation and simplify, giving $$\overline{AB'} = r^2 + x^2 + r x \sqrt{2 - 2 \cos\alpha}.$$ Then $AB'$ and $AB''$ in your figure can be calculated by setting $x=h$ and $x=2h$.