Angles

Geometry Level 3

The figure below is made on the regular grid.

Find the value of ${\color{#3D99F6}{\text{blue angle}}} \ - \ {\color{#D61F06}{\text{red angle}}}.$

45° 30° None of the others 60°

2 solutions

Áron Bán-Szabó
Aug 21, 2017

$\triangle ABC$ is isosceles, because $AC=BC$ . In the other hand $\angle ACB=90°$ , so $\angle BAC=\angle CBA = 45°$ . (If you don't know why, just ask in comment, and I complete my solution.) Since $\angle DAC={\color{#3D99F6}{\text{blue angle}}}$ , ${\color{#3D99F6}{\text{blue angle}}} \ - \ {\color{#D61F06}{\text{red angle}}}=\boxed{45°}.$

I find it troublesome that the problem is given in a 2 by 6 matrix, the solution is given in a 3 by 6 matrix, the node C is not in the same place in the problem and solution. In the solution diagram the blue angle is clearly greater than 45 degrees.In the problem drawing, it seems that the blue angle is less than 45 degrees.Ed Gray

Edwin Gray - 3 years, 9 months ago

Dick van der Leeden
Apr 14, 2019

$\sin(blue)=\dfrac{3}{\sqrt{13}}$

$\cos(blue)=\dfrac{2}{\sqrt{13}}$

$\sin(red)=\dfrac{1}{\sqrt{26}}$

$\cos(red)=\dfrac{5}{\sqrt{26}}$

$\sin(blue-red)=\dfrac{3}{\sqrt{13}}\cdot\dfrac{5}{\sqrt{26}}-\dfrac{2}{\sqrt{13}}\cdot \dfrac{1}{\sqrt{26}} = \dfrac{1}{\sqrt{2}}\rightarrow blue - red = 45^o$

Angles

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