In a triangle PQR, side PQ=3, and side PR=2cm. The area of the triangle PQR is 2cm^2.
State two possible values for angle QPR (in degrees).
+128405 Here's one way to do this :
Let P = (0,0)
Let Q = (3,0)
Connstruct a circle centered at P with a radius of 2
The equation is x^2 + y ^2 = 4
Let the base of PQR = PQ = 3
So....the height of PQR can be found as
2 = (1/2) 3 * height
2 = 3/2 * height
height = (2/3)(2) =4/3
Sub this into the equation of the circle for y
x^2 + (4/3)^2 = 4
x^2 + 16/9 = 4
x^2 = (36 - 16) /9
x^2 = 20 / 9 take both roots
x = sqrt (20)/3 = (2/3) sqrt (5)
x = - sqrt(20)/3 = (-2/3)sqrt (5)
So ....as shown in the figure
R1 = ( -(2/3)sqrt 5 ,4/3)
R2 = ( (2/3)sqrt (5) ,4/3)
Possible angle values of QPR ≈ 41.81° and ≈ 138.19°
