Answer: ![m\angle C=53\°](https://tex.z-dn.net/?f=m%5Cangle%20C%3D53%5C%C2%B0)
Step-by-step explanation:
For this exercise you need to use the Inverse Trigonometric function arcsine, which is defined as the inverse function of the sine.
Then, to find an angle α, this is:
![\alpha =arcsin(\frac{opposite}{hypotenuse})](https://tex.z-dn.net/?f=%5Calpha%20%3Darcsin%28%5Cfrac%7Bopposite%7D%7Bhypotenuse%7D%29)
In this case, you can identify that:
![\alpha =m\angle C\\\\opposite=AB=36\ cm\\\\hypotenuse=AC=45\ cm](https://tex.z-dn.net/?f=%5Calpha%20%3Dm%5Cangle%20C%5C%5C%5C%5Copposite%3DAB%3D36%5C%20cm%5C%5C%5C%5Chypotenuse%3DAC%3D45%5C%20cm)
Then, substituting values into
and evaluating, you get that the measure of the angle "C" to the nearest degree, is:
![m\angle C=arcsin(\frac{36\ cm}{45\ cm})\\\\m\angle C=53\°](https://tex.z-dn.net/?f=m%5Cangle%20C%3Darcsin%28%5Cfrac%7B36%5C%20cm%7D%7B45%5C%20cm%7D%29%5C%5C%5C%5Cm%5Cangle%20C%3D53%5C%C2%B0)
Answer:
<h3><u>Let's</u><u> </u><u>understand the concept</u><u>:</u><u>-</u></h3>
Here angle B is 90°
So
and
Are right angled triangle
So we use Pythagoras thereon for solution
<h3><u>Required Answer</u><u>:</u><u>-</u></h3>
perpendicular=p=8cm
Hypontenuse =h =10cm
According to Pythagoras thereon
![{\boxed{\sf b^2=h^2-p^2}}](https://tex.z-dn.net/?f=%7B%5Cboxed%7B%5Csf%20b%5E2%3Dh%5E2-p%5E2%7D%7D)
![\longrightarrow](https://tex.z-dn.net/?f=%5Clongrightarrow)
![\sf b^2=10^2-p^2](https://tex.z-dn.net/?f=%5Csf%20b%5E2%3D10%5E2-p%5E2)
![\longrightarrow](https://tex.z-dn.net/?f=%5Clongrightarrow)
![\sf b={\sqrt {10^2-8^2}}](https://tex.z-dn.net/?f=%5Csf%20b%3D%7B%5Csqrt%20%7B10%5E2-8%5E2%7D%7D)
![\longrightarrow](https://tex.z-dn.net/?f=%5Clongrightarrow)
![\sf b={\sqrt{100-64}}](https://tex.z-dn.net/?f=%5Csf%20b%3D%7B%5Csqrt%7B100-64%7D%7D)
![\longrightarrow](https://tex.z-dn.net/?f=%5Clongrightarrow)
![\bf b={\sqrt {36}}](https://tex.z-dn.net/?f=%5Cbf%20b%3D%7B%5Csqrt%20%7B36%7D%7D)
![\longrightarrow](https://tex.z-dn.net/?f=%5Clongrightarrow)
![\sf b=6](https://tex.z-dn.net/?f=%5Csf%20b%3D6)
![\therefore](https://tex.z-dn.net/?f=%5Ctherefore)
![\overline{BC}=6cm](https://tex.z-dn.net/?f=%5Coverline%7BBC%7D%3D6cm)
![\longrightarrow](https://tex.z-dn.net/?f=%5Clongrightarrow)
![BD=9+6](https://tex.z-dn.net/?f=BD%3D9%2B6)
![\longrightarrow](https://tex.z-dn.net/?f=%5Clongrightarrow)
![BD=15cm](https://tex.z-dn.net/?f=BD%3D15cm%20)
- Now in
![\triangle ABD](https://tex.z-dn.net/?f=%5Ctriangle%20ABD%20)
Perpendicular=p=8cm
Base =b=15cm
- We need to find Hypontenuse =AD(x)
According to Pythagoras thereon
![{\boxed {\sf h^2=p^2+b^2}}](https://tex.z-dn.net/?f=%7B%5Cboxed%20%7B%5Csf%20h%5E2%3Dp%5E2%2Bb%5E2%7D%7D)
![\longrightarrow](https://tex.z-dn.net/?f=%5Clongrightarrow)
![\sf h^2=8^2+15^2](https://tex.z-dn.net/?f=%5Csf%20h%5E2%3D8%5E2%2B15%5E2%20)
![\longrightarrow](https://tex.z-dn.net/?f=%5Clongrightarrow)
![\sf h={\sqrt {8^2+15^2}}](https://tex.z-dn.net/?f=%5Csf%20h%3D%7B%5Csqrt%20%7B8%5E2%2B15%5E2%7D%7D)
![\longrightarrow](https://tex.z-dn.net/?f=%5Clongrightarrow)
![\sf h={\sqrt {64+225}}](https://tex.z-dn.net/?f=%5Csf%20h%3D%7B%5Csqrt%20%7B64%2B225%7D%7D)
![\longrightarrow](https://tex.z-dn.net/?f=%5Clongrightarrow)
![\sf h={\sqrt {289}}](https://tex.z-dn.net/?f=%5Csf%20h%3D%7B%5Csqrt%20%7B289%7D%7D)
![\longrightarrow](https://tex.z-dn.net/?f=%5Clongrightarrow)
![\sf h=17cm](https://tex.z-dn.net/?f=%5Csf%20h%3D17cm%20)
![\therefore](https://tex.z-dn.net/?f=%5Ctherefore)
![{\underline{\boxed{\bf x=17cm}}}](https://tex.z-dn.net/?f=%7B%5Cunderline%7B%5Cboxed%7B%5Cbf%20x%3D17cm%7D%7D%7D)
Answer:
1
Step-by-step explanation:
you simply divide 154 by 9, get around 17.1. 17 is the closest whole number answer, so 17*9 is 153.
do 154-153, you get 1.
not the best explaination but i hope it helps.
Hello, Julio can be no younger then 11. Hoped I helped!