Question #179497

Other Math

Given the following data of a quadrilateral: ∠ABC=104°, ∠BCD=89°, ∠ABD=60°, ∠ACD=53°, and AD=1000 m

1.)Determine the length of line BC (in meters, 2 decimal place):   

2.)Determine ∠CAD:

3.)Determine ∠BDA:



1
Expert's answer
2021-05-07T09:59:34-0400


Values to find: x,y,BCx, y, BC (see picture).

1) Obviously x+y=80°x+y=80\degree. Via Sinus Theorem

AB=ACsin(36°)sin(104°)AB=\frac{AC \sin(36\degree)}{\sin(104\degree)} & AD=ACsin(53°)sin(37°+y)AD=\frac{AC\sin(53\degree)}{\sin(37\degree+y)} & AB=ADsin(y)sin(60°)AB=\frac{AD\sin(y)}{\sin(60\degree)} =>

ACsin(53°)sin(y)sin(37°+y)sin(60°)=ACsin(36°)sin(104°)\frac{AC\sin(53\degree)\sin(y)}{\sin(37\degree+y)\sin(60\degree)}=\frac{AC\sin(36\degree)}{\sin(104\degree)}

Using sin(53°)=sin(90°37°)=cos(37°)\sin(53\degree)=\sin(90\degree-37\degree)=\cos(37\degree) we can rewrite previous equation as

cos(37°)sin(y)sin(37°)cos(y)+cos(37°)sin(y)=sin(60°)sin(36°)sin(104°)\frac{\cos(37\degree)\sin(y)}{\sin(37\degree)\cos(y)+\cos(37\degree)\sin(y)}=\frac{\sin(60\degree)\sin(36\degree)}{\sin(104\degree)} =>

sin(104°)sin(60°)sin(36°)=1+tg(37°)ctg(y)\frac{\sin(104\degree)}{\sin(60\degree)\sin(36\degree)}=1+\tg(37\degree)\ctg(y) =>

y=arctg[1tg(37°)(sin(104°)sin(60°)sin(36°)1)]y=\arctg\biggr[\frac{1}{\tg(37\degree)}(\frac{\sin(104\degree)}{\sin(60\degree)\sin(36\degree)}-1)\biggl]

yarctg(1,19)40°y \approx\arctg(1,19)\approx 40\degree

x=80°y40°x=80\degree-y\approx 40\degree

2) Via Sinus Theorem AB=BCsin(36°)sin(40°)AB=\frac{BC\sin(36\degree)}{\sin(40\degree)} & AB=ADsin(y)sin(60°)AB=\frac{AD\sin(y)}{\sin(60\degree)} =>

BC=ADsin(y)sin(40°)sin(60°)sin(36°)BC=\frac{AD\sin(y)\sin(40\degree)}{\sin(60\degree)\sin(36\degree)}

BC1000m0,798=798mBC\approx 1000m*0,798=798m


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