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enter image description hereCalculate $\alpha$ if $BC=AD$

(Answer:$10^o $)

I have the resolution below but I would like a geometric resolution by auxiliary lines

Sine Theorem in $\triangle{BCD}$:

$$\frac{BC}{BD}=\frac{\sin(180−3\theta)}{\sin(2\theta)}=\frac{\sin(3\theta)}{\sin(2\theta)}$$

Sine Theorem in $\triangle{BDA}$:

$$\frac{BC}{BD}=\frac{AD}{BD}=\frac{\sin(180−7\theta)}{\sin(4\theta)}=\frac{\sin(7\theta)}{\sin(4\theta)}=\frac{\sin(7\theta)}{2\sin(2\theta)\cos(2\theta)}$$

$$\angle{ABD}=180−4\theta−\angle{BDA}=180$$

Sine Theorem in $\triangle{BDA}$:

$$\theta−(180−\angle{BDC})=180−4\theta−3\theta=180−7\theta$$

Equating both equations:

$$2\sin(3\theta)\cos(2\theta)=\sin(7\theta)$$

$$\sin(5\theta)+\sin(\theta)=\sin(7\theta)$$

$$\sin(\theta)=\sin(7\theta)−\sin(5\theta)$$

$$\sin(\theta)=2\cos(6\theta)\sin(\theta)$$

$$\cos(6\theta)=1/2⇒6\theta=60⇒θ=10$$

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  • $\begingroup$ No cosine rule right? $\endgroup$ Commented Mar 17 at 12:44
  • $\begingroup$ @AnuradhaBanthia Yes....The question belongs to a geometry book where trigonometric formulas are hardly used, but auxiliary constructions $\endgroup$ Commented Mar 17 at 13:25
  • $\begingroup$ i have 2 solution but both are trig $\endgroup$ Commented Mar 17 at 13:37
  • $\begingroup$ @AnuradhaBanthia But post anyway, maybe others would like to know $\endgroup$ Commented Mar 17 at 13:49
  • $\begingroup$ @AnuradhaBanthia The answer is correct $\endgroup$ Commented Mar 17 at 15:01

2 Answers 2

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► Draw the angles $\angle{B'AB}=2a$ and $\angle{B'CB}=4a$ so we get an isosceles triangle $\triangle{B'AC}$ and $\angle{B'AC}=\angle{B'CA}=6a$.

►► Draw the circle centered at $A$ with radius $AC$. The intersection of lines $AB'$ and $CB'$ is the point $B''=\left(\dfrac{AC}{2},\dfrac{\tan(6a)AC}{2}\right)$. $B''$ is in this circle if and only if $B''=B'$ so $\triangle {B'AC}$ is equilateral because $AB'=B'C$ (equal sides) and $AC=AB'$ (both equal to the radius).

►►► If $x^2+y^2=AC^2$ then $\left(\dfrac {AC}{2}\right)^2+\left(\dfrac{AC\tan(6a)}{2}\right)^2=AC^2\iff\left(\dfrac {1}{2}\right)^2+\left(\dfrac{\tan(6a)}{2}\right)^2=1$ This implies $6a=60^{\circ}$ so $a=10^{\circ}$

enter image description here

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I cannot seem to find an easier solution by auxiliary lines. Although I have $2-3$ approaches to make it easy.

Now we extend perpendicular from $B $ to $AC$ at $E$. Assume alpha is $ a $ in my text

Let $BE=x$.

Hence $ x \space \cot 4a$ $ + x\space \cot 3a =x \space \csc 2a$.

So $ \space \cot 4a$ $ + \space \cot 3a =\space \csc 2a$.

Now this might seem like a tough equation but plotting them on a graph makes it very easy if u know the curves of the equations. (This can be solved without graphs also). Upon solving u get $a=10 ^{\circ}$

I think the other way is to use cosine rule with $a$ which has little more bash.

U can try by coordinate geometry also by assuming $A$ to be your $(0,0)$ and solve with angles and slopes.

This is the diagram sorry if i messed up the labels as I have done points as your figure. enter image description here

I hope this helps.

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  • $\begingroup$ Pretty cool suggestion, I might check if the solution is easier with coordinate geometry later $\endgroup$ Commented Mar 17 at 15:14
  • $\begingroup$ U could also assume BC as x-axis and AE as y-axis and E to be your origin instead to make it easier perhaps $\endgroup$ Commented Mar 17 at 15:15

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