Question 455021: The sides of a triangular plot of land are 50, 80, and 100 meters.
a) Find to the nearest degree the measure of the largest angle of the triangle.
b) Using the answer obtained in part a, find the area of the triangle to the nearest square meter.
Answer by lwsshak3(11628)   (Show Source):
You can put this solution on YOUR website! The sides of a triangular plot of land are 50, 80, and 100 meters.
a) Find to the nearest degree the measure of the largest angle of the triangle.
b) Using the answer obtained in part a, find the area of the triangle to the nearest square meter
..
Draw a triangle as follows:
A=apex
B=right angle
C=left angle
D=point where vertical line from apex intersects with BC
CA=50m, BC=100m (base), AB=80m
..
a) Using Law of cosines to find largest angle:(opposite BC)
100^2=80^2+50^2-2*80*50 Cos A
10000=6400+2500-8000 Cos A
1100=-8000 cos A
Cos A=1100/-8000=-.1375
A=97.90º
..
b) Using Law of sines
sin A/100=sin C/50
sin C=(50/100)sin A=sin 97.9/2=.4953
C=29.69º
..
Sin C=AD/80
AD=80 sin C=80 sin 29.69=39.62(height of triangle)
Area=1/2 Base*height=1/2*100*39.62=1981 m^2
..
ans:
The largest angle of the triangle=98º
Area of the triangle=1981 m^2
|
|
|
