since PQ > PR sin50°, angle Q can be either acute or obtuse.
sinQ/11 = sin50°/9
there are two angles Q which satisfy that equation.
In triangle PQR, angle PRQ= 50°, PR=11, and PQ=9
a) Show that there are two possible measures of angle PQR.
b) Sketch triangle PQR for each case.
c) For each case, find: i) the measure of angle QPR, ii) the area of the triangle, iii) the perimeter of the triangle.
pls help
2 answers
Part a:
Sin PQR/ 11 = Sin 50°/9
Sin PQR = 11sin50°/9
PQR = sin^-1(11sin50°/9)
That would give you around 69.4°(Q1)
180° - 69.4° ~ 110.6°(Q2)
Part b: express the acute answer (Q1) and obtuse (Q2) on the triangle
Part C.i.: split the entire triangle in half to make 2 right triangles
Left side right angle: 11cos50° ~ 7.07
Right side right angle: 9sin20° ~ 3.08
Height of both triangles: 11sin50 ~ 8.43
Left side right angle of P: cos^-1(8.43/11) ~ 40.0°
Right side of right triangle: cos^1(8.43/9) ~ 20.6°
40.0 + 20.6 = PQR = 60.6
Still trying to figure out C.ii. and c.iii
Sin PQR/ 11 = Sin 50°/9
Sin PQR = 11sin50°/9
PQR = sin^-1(11sin50°/9)
That would give you around 69.4°(Q1)
180° - 69.4° ~ 110.6°(Q2)
Part b: express the acute answer (Q1) and obtuse (Q2) on the triangle
Part C.i.: split the entire triangle in half to make 2 right triangles
Left side right angle: 11cos50° ~ 7.07
Right side right angle: 9sin20° ~ 3.08
Height of both triangles: 11sin50 ~ 8.43
Left side right angle of P: cos^-1(8.43/11) ~ 40.0°
Right side of right triangle: cos^1(8.43/9) ~ 20.6°
40.0 + 20.6 = PQR = 60.6
Still trying to figure out C.ii. and c.iii
2 answers