A quadrilateral ABCD in which angle A is 90°, AD=6cm,angle ADB =40°,angle DBC =5° and angle DCB =45°. Then find the area of quadrilateral ABCD .

							
Draw a quadrilateral ABCD.join AD.Look that AD divides the quadrilateral in two triangles.Now follow the steps I say.
Remember the formula,sum of all three angles of a triangle is 180°.
 
In the triangle ABD,angle DAB is 90°  and angle ADB is 40° given.So angle DBA=180-(90+40)=50°.
In triangle BCD, angle DBC=5 and angle DCB=45.So angle BCD=180-(5+45)=130°.
Now we have to find out area of these two triangke and then add up.
In triangle ABD it is simple.
Area ABD=(1/2)AB×AD=(1/2)AB×6=3AB.
We have to find out AB.
See tan50°=6/AB
or,AB=6cot50°.
So area ABD=18cot50°.
For triangle BCD,drop a perpendicular from C on BD and name the point where it touches BD as Q.So angle CQB=90°.This CQ is the height of triangle BCD and BD is the base.
We can find out BD using triangle ABD as
sin50°=AD/BD
or,BD=6cosec50°.
From triangle BCD,using sin law
sin45°/BD=sin130°/BC
or,BC=BDsin130°/sin45°=BD(root2)sin(180-50)=BD(root2)sin50°.
From triangle CQB,
sin5°=CQ/BC
or,CQ=BCsin5°=BD(root2)sin50°sin5°=6(root2)cosec50sin50sin5=6(root2)sin5.
So area of BCD is
(1/2)×BD×CQ=(1/2)×6cosec50×6(root2)sin5=18(root2)sin5cosec50.
Add this two areas and that is the area of the quadrilateral.
Please approve the answer if helped.