Answer:
-9/10
Step-by-step explanation:
negative nine tenths
Recall that
![\tan(\alpha-\beta)=\dfrac{\tan\alpha-\tan\beta}{1+\tan\alpha\tan\beta}](https://tex.z-dn.net/?f=%5Ctan%28%5Calpha-%5Cbeta%29%3D%5Cdfrac%7B%5Ctan%5Calpha-%5Ctan%5Cbeta%7D%7B1%2B%5Ctan%5Calpha%5Ctan%5Cbeta%7D)
![\cos^2\theta+\sin^2\theta=1](https://tex.z-dn.net/?f=%5Ccos%5E2%5Ctheta%2B%5Csin%5E2%5Ctheta%3D1)
From the second identity, we find
![\cos x=\pm\sqrt{1-\sin^2x}](https://tex.z-dn.net/?f=%5Ccos%20x%3D%5Cpm%5Csqrt%7B1-%5Csin%5E2x%7D)
![\sin y=\pm\sqrt{1-\cos^2y}](https://tex.z-dn.net/?f=%5Csin%20y%3D%5Cpm%5Csqrt%7B1-%5Ccos%5E2y%7D)
We're told that
and
, which means we should take the positive roots above. Then
![\cos x=\sqrt{1-\left(\dfrac8{17}\right)^2}=\dfrac{15}{17}](https://tex.z-dn.net/?f=%5Ccos%20x%3D%5Csqrt%7B1-%5Cleft%28%5Cdfrac8%7B17%7D%5Cright%29%5E2%7D%3D%5Cdfrac%7B15%7D%7B17%7D)
![\sin y=\sqrt{1-\left(\dfrac35\right)^2}=\dfrac45](https://tex.z-dn.net/?f=%5Csin%20y%3D%5Csqrt%7B1-%5Cleft%28%5Cdfrac35%5Cright%29%5E2%7D%3D%5Cdfrac45)
Then we compute the tangents:
![\tan x=\dfrac{\sin x}{\cos x}=\dfrac{\frac8{17}}{\frac{15}{17}}=\dfrac8{15}](https://tex.z-dn.net/?f=%5Ctan%20x%3D%5Cdfrac%7B%5Csin%20x%7D%7B%5Ccos%20x%7D%3D%5Cdfrac%7B%5Cfrac8%7B17%7D%7D%7B%5Cfrac%7B15%7D%7B17%7D%7D%3D%5Cdfrac8%7B15%7D)
![\tan y=\dfrac{\sin y}{\cos y}=\dfrac{\frac45}{\frac35}=\dfrac43](https://tex.z-dn.net/?f=%5Ctan%20y%3D%5Cdfrac%7B%5Csin%20y%7D%7B%5Ccos%20y%7D%3D%5Cdfrac%7B%5Cfrac45%7D%7B%5Cfrac35%7D%3D%5Cdfrac43)
so we end up with
![\tan(x-y)=\dfrac{\frac8{15}-\frac43}{1+\frac8{15}\cdot\frac43}=-\dfrac{36}{77}](https://tex.z-dn.net/?f=%5Ctan%28x-y%29%3D%5Cdfrac%7B%5Cfrac8%7B15%7D-%5Cfrac43%7D%7B1%2B%5Cfrac8%7B15%7D%5Ccdot%5Cfrac43%7D%3D-%5Cdfrac%7B36%7D%7B77%7D)
Answer:
Since AC is perfectly equal to BD, 4x-60=30-x. If you solve for x, x=18. Then plug 18 into BD=30-x. Therefore, BD=12.
Answer:
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