SOLUTION: In triangle PQR, P = 50°, PR = 11 and PQ = 9. a) Show that there are two possible measures of PQR b) Sketch triangle PQR for each case c)For each case, find: i) the meas

Algebra ->  Trigonometry-basics -> SOLUTION: In triangle PQR, P = 50°, PR = 11 and PQ = 9. a) Show that there are two possible measures of PQR b) Sketch triangle PQR for each case c)For each case, find: i) the meas      Log On


   

Question 1182314: In triangle PQR, P = 50°, PR = 11 and PQ = 9.
a) Show that there are two possible measures of PQR
b) Sketch triangle PQR for each case
c)For each case, find: i) the measure of QPR , ii) the area of the triangle, iii) the perimeter of the triangle.
Found 2 solutions by math_tutor2020, MathTherapy:
Answer by math_tutor2020(3817) About Me  (Show Source):
You can put this solution on YOUR website!

Only one triangle is possible because of the SAS congruence rule.

Let's say that triangle ABC and triangle PQR had the following measurements
  • angle A = 50, angle P = 50
  • AC = 11, PR = 11
  • AB = 9, PQ = 9
For triangle ABC, the angle A is between the mentioned sides AB and AC
For triangle PQR, the angle P is between the mentioned sides PR and PQ

Diagram:

In other words, if you know two sides of a triangle, and the included angle, then only one triangle is possible. The diagram above shows that if you can claim two different triangles are possible, then you arrive at a contradiction due to the SAS rule.

So I'm not sure why your teacher thinks there are two triangles possible. There may be a typo somewhere.

Answer by MathTherapy(10555) About Me  (Show Source):
You can put this solution on YOUR website!
In triangle PQR, P = 50°, PR = 11 and PQ = 9.
a) Show that there are two possible measures of PQR
b) Sketch triangle PQR for each case
c)For each case, find: i) the measure of QPR , ii) the area of the triangle, iii) the perimeter of the triangle.
Is measure of PQR referring to ∡PQR? If so, don't you think you need to state that? 

If so and measure of QPR refers to ∡QPR, which is the same as ∡P, wasn't that given?

a) Use the 2 given sides, the included angle, and Law of Cosines to find side p (same as side QR).

   This should be around 8.644539 units. Then use the Law of Sines to find ∡Q, which should be approximately 78.47o.   

   With ∡s P and Q being 50o, and 78.47o, respectively, ∡R is then 51.53o.

   Note that ∡Q, being 78.47o can also measure 101.53o since its reference angle measures that, in the 2nd quadrant.

   With ∡Q being 101.53o, ∡P, 50o, then ∡R becomes 28.47o. This proves that ∡Q (same as ∡PQR) can have 2 measures. 

   This also means that ∆PQR can either be ACUTE or OBTUSE.

   Note that with 2 sides and an INCLUDED angle given, the requested AREA of this NON-RIGHT triangle can be found

   by using the formula: highlight_green%28matrix%281%2C3%2C+Area%2C+%22=%22%2C+%281%2F2%29ab%2ASin+%28C%29%29%29, which in this case would be: highlight_green%28matrix%281%2C3%2C+Area%2C+%22=%22%2C+%281%2F2%29qr%2Asin+%28P%29%29%29

   You should now be able to answer the other questions.