Trig + Triangle Laws: How come?

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Trig + Triangle Laws: How come?

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#1 h Mar 29, 2022, 8:26 PM

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My question is after the problem and my solution.
Problem:
In the quadrilateral $ABCD$ , $\angle{ADC}=90^\circ$ , $AB=2$ , $BD=5$ .
(1) Find $\cos{\angle{ADB}}$ .
(2) If $DC=2\sqrt{2}$ , find $BC$ . (2018 China Gaokao Syllabus I #17)

Solution:
Draw a diagram (see the attachment):
(1) First, $\frac{5}{\sin{A}}=\frac{2}{\sin{\angle{ADB}}}$ . Since $\sin{A}=\frac{\sqrt{2}}{2}$ , we can plug it back and find that $\sin{\angle{ADB}}=\frac{\sqrt{2}}{5}$ . Therefore, through the Pythagorean Identities, we find that $\cos{\angle{ADB}}=\boxed{\frac{\sqrt{23}}{5}}$ .

(2) Notice that since $\angle{ADC}=90^\circ$ , its sine is $1$ . Therefore, using the fact that $\angle{ADC}=\angle{ADB}+\angle{BDC}$ , we can say: $\sin{\angle{ADC}}=\sin{(\angle{ADB}+\angle{BDC})}=1$ .
Expand using sum of sines: $\frac{\sqrt{2}}{5}\cos{\angle{BDC}}+\frac{\sqrt{23}}{5}\sin{\angle{BDC}}=1$ .
Using $(\sin{\angle{BDC}})^2+(\cos{\angle{BDC}})^2=1$ , call $\cos{\angle{BDC}}$ equal to $x$ (notice that since $\angle{BDC}$ is acute, its sine must be positive, so that's why I used the cosine to set the value to avoid problems), $\sin{\angle{BDC}}=\sqrt{1-x^2}$ . Plug it back in and solve the quadratic equation, we find that $c=\frac{\sqrt{2}}{5}$ . Now using the law of cosines, we find that $BC=\sqrt{5^2+(2\sqrt{2})^2-2*5*2\sqrt{2}*\frac{\sqrt{2}}{5}}=\boxed{5}$ .

Wow! Such a long bash! Thanks for reading the solution. Now, here's the question:
In the answer key to part (2) of this problem, it said:

Because of the problem's given information and part (1) [its solution is identical to mine, just differently worded], it is obvious that $\cos{\angle{BDC}}=\sin{\angle{ADB}}$ .

How come? I bashed all of this out to obtain that fact. How come it's "obvious?" Please explain. Notice that if the answer key omitted these details, it wouldn't be an answer key because the bashing is vital to this problem it looks like!

Thanks!

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#2 h Mar 29, 2022, 8:28 PM

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I used GeoGebra since I didn't know how to create an image using LaTeX. Can someone give me help on that? Sorry. I read the wiki and tutorials and I'm still very stuck on how to create images in LaTeX.

Thanks!

This post has been edited 1 time. Last edited by hastapasta, Mar 29, 2022, 8:46 PM

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#3 h Mar 29, 2022, 9:03 PM

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Notice that $\angle BDC+\angle ADB=90^\circ.$ So the fact that $\cos\angle BDC=\sin \angle ADB$ is a direct result from the fact that $\cos\theta=\sin\left(90^\circ-\theta\right).$ The proof of this isn't too hard. Just consider the unit circle. Notice that rotating the point $(1,0)$ by an angle of $\theta$ counterclockwise takes it to the point $(\cos\theta,\sin\theta),$ and rotating it $\theta$ clockwise sends it to $(\cos(90^\circ-\theta),\sin(90^\circ-\theta)).$ Also notice that these two points are symmetric about $y=x,$ so the points $\cos(90^\circ-\theta),\sin(90^\circ-\theta))$ and $(\sin\theta,\cos\theta)$ are the same, so $\sin\theta=\cos(90^\circ-\theta)$ and $\cos\theta=\sin(90^\circ-\theta).$

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#4 h Mar 29, 2022, 9:35 PM

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Oh, yeah! Didn't realize that! Thanks!

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Carlos Jacob Rubio Barrio

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Carlos Jacob Rubio Barrio

#5 h Mar 28, 2023, 2:23 PM

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Why sin A=\sqrt{2}/2???

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