Answer to Question #179497 in Geometry for Nikita Ramesh Gaikwad

Question

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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:

Answer 1

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

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

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

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

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

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

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

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

$y \approx\arctg(1,19)\approx 40\degree$

$x=80\degree-y\approx 40\degree$

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

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

$BC\approx 1000m*0,798=798m$