How to find angle inside a right isosceles triangle with one side and an angle given

Question

The problem is as follows:

The picture from below shows a triangle. The conditions are stated in the diagram.

The alternatives given in my book are as follows:

$\begin{array}{ll} 1.&30^{\circ}\\ 2.&15^{\circ}\\ 3.&25^{\circ}\\ 4.&18^{\circ}\\ \end{array}$

I'm not sure exactly what to do with this problem. So far the only thing which I could spot on was that the right triangle is $45-45-90$ but other than that I'm stuck. Can somebody guide me in the right direction with this problem?.

Answer 1

Let ∠DAC = ∠DCB = $x$ and ∠DCA = ∠DAB = $y$. Then, $x+y = 45$ due to the isosceles right triangle ABC and ∠ADC = 180 - ($x+y$) = 135. Apply the sine rule to the triangle ADC,

$$\frac{\sin x}{\sin ∠ADC} = \frac{DC}{AC} = \frac1{\sqrt2}$$

which leads to $\sin x = \frac1{\sqrt2}\cdot \sin 135 = \frac 12$. Thus, ∠DCB = $x$ = 30.

Answer 2

Here is a solution using only synthetic geometry. The following figure is the "solution by picture":

$D$ is on the two emphasized circles, $\Delta XCD$ is equilateral, so $\widehat{DCB}=30^\circ$.

$\square$

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Details: (If really needed.) Let $X$, be the fourth point of the square with vertices $A,B,C$.

• The given conditions conditions describe an isosceles triangle $\Delta ABC$ with a right angle in $B$,

• and the point $\color{brown}D$ is at the intersection

  • of the circle $(C)$ centered in $C$ through $B$ (and $X$), since $CB=CD$,
  • with the circle $(X)$ centered in $X$ through $A,C$. This is because $\widehat{DAC}=\widehat{BCD}$ implies $$ \begin{aligned} \widehat{ADC}&=180^\circ-\widehat{DAC}-\widehat{DAC} \\&=180^\circ-\widehat{BCD}-\widehat{DAC} \\&=180^\circ-\widehat{BCA} \\&=180^\circ-45^\circ=135^\circ\ . \end{aligned}$$ (A point on the (small) arc $\overset \frown{AC}$ of the circle $(X)$ has the property that $\widehat{ADC}$ has half of the measure of the (big) arc arc $\overset \frown{CA}$, with measure $360^\circ-\widehat{AXC} = 360^\circ-90^\circ=270^\circ$. So it measures $135^\circ$.)

Because of $CD=DX=XC$, the triangle $\Delta CDX$ is equilateral, we obtain an angle of $60^\circ$ in $C$, so $$ \widehat{DCB}= \widehat{XCB}- \widehat{XCD}= 90^\circ-60^\circ=30^\circ\ . $$

$\square$

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Bonus: Let us also draw the other quarter circles centered in the other vertices of the square $ABCX$, each having radius $AB=BC=CX=XA$. By the symmetry of the picture, we obtain as in the picture intersections $D,E,F,G$, which are forming a square. With the arguments above,

• $ABCX$, $DEFG$ are squares,
• and the triangles $\Delta CDX$, $\Delta XEA$, $\Delta AFB$, $\Delta BGC$; $\Delta AGD$, $\Delta BDE$, $\Delta CEF$, $\Delta XFG$ are equilateral.

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Later EDIT:

A simplified picture was added:

(This picture is suited for a straightforward proof for the posted problem, but it hides the symmetry in the picture. Note that many such problems arise as particular, partial pictures in a constellation involving a regular polygon. Finding it is a good way to understand the special situation, and to be able to react in similar situations. Or to even compose problems.)