Find the area of triangle ABC:
AB=19, ∠B= 32°.
A line is drawn from C down to AB. We call the point of intersection D.
AD= 5, ∠ACD=10°.
I don't know how to set up the equations to find x and y. Help is much appreciate
+33614 See if this helps:
(1) 10° + 32° + θ + Φ = 180°
(2) sin(32°)/CD = sin(θ)/12
(3) sin(Φ)/CD = sin(10°)/5
Three equations in three unknowns should enable you to find the angles.
Then use the sin rule again to find the lengths of the other sides, AC and BC.
Then use Herons formula to find the area:
Area = √[s(s - AB)(s - AC)(s - BC)] where s = (AB + BC + AC)/2
.
+33614 See if this helps:
(1) 10° + 32° + θ + Φ = 180°
(2) sin(32°)/CD = sin(θ)/12
(3) sin(Φ)/CD = sin(10°)/5
Three equations in three unknowns should enable you to find the angles.
Then use the sin rule again to find the lengths of the other sides, AC and BC.
Then use Herons formula to find the area:
Area = √[s(s - AB)(s - AC)(s - BC)] where s = (AB + BC + AC)/2
.
