For each of the following, tell how many non congruent triangles PQR fit the given description, and find the size of angle Q. Make a separate diagram for each case. (a) p = 3, q = 5, angle P = 27 degrees (b) p = 8, q = 5, angle P = 57 degrees (c) p = 7, q = 8, angle P = 70 degrees (d) p = 10, q = 20, angle P = 30 degrees.
Oh, triangles! My favorite shape sideshow! Let's get cracking on making some diagrams and solving these puzzles.
(a) p = 3, q = 5, angle P = 27 degrees:
To create non-congruent triangles, we need to fix the lengths of the sides and only play around with angles. Since the angle is already specified, we can only have one triangle that fits this description. Here's a diagram for you:
```
P
/|
/ |
q / | p
/ |
/ |
/ |
R /______|
q + p + q
```
Angle Q is the remaining angle, which makes it 180 - 27 - 90 = 63 degrees.
(b) p = 8, q = 5, angle P = 57 degrees:
Let me whip up a diagram for you:
```
P
/|
/ |
q / | p
/ |
/ |
R / |
/______|
q + p + q
```
Since angle P is specified again, we only have one unique triangle here. To find angle Q: 180 - 57 - 90 = 33 degrees.
(c) p = 7, q = 8, angle P = 70 degrees:
Here's a triangle-tastic diagram for you:
```
P
/|
/ |
q / | p
/ |
/ |
R / |
/______|
q + p + q
```
We still have one non-congruent triangle here. To determine angle Q: 180 - 70 - 90 = 20 degrees.
(d) p = 10, q = 20, angle P = 30 degrees:
Buckle up for another splendid diagram:
```
P
/|
/ |
q / | p
/ |
/ |
R / |
/______|
q + p + q
```
This time, we still only have one non-congruent triangle. To calculate angle Q: 180 - 30 - 90 = 60 degrees.
Now you have the scoop on the number of non-congruent triangles, as well as the size of angle Q for each case. Enjoy your triangle escapades!
(a) To find the number of non-congruent triangles PQR that fit the given description, use the Triangle Inequality Theorem. According to this theorem, the sum of the lengths of any two sides of a triangle must be greater than the length of the third side.
In this case, p = 3 and q = 5. Using the Triangle Inequality Theorem, we have:
3 + 5 > x,
8 > x.
Therefore, the possible range of length for side r is 0 < r < 8.
To find the size of angle Q, use the Law of Cosines:
p^2 = q^2 + r^2 - 2qr*cos(P).
Substituting the given values:
3^2 = 5^2 + r^2 - 2*5*r*cos(27).
9 = 25 + r^2 - 10r*cos(27).
r^2 - 10r*cos(27) + 16 = 0.
To solve this quadratic equation, substitute the values of cos(27) and simplify:
r^2 - 8.67r + 16 = 0.
Using the quadratic formula, we find:
r ≈ 5.656 or r ≈ 2.813.
Since 0 < r < 8, the only valid value is r ≈ 2.813.
To find angle Q, use the Law of Sines:
sin(Q)/q = sin(P)/p.
Substituting the known values:
sin(Q)/5 = sin(27)/3.
Cross-multiply:
sin(Q) = (5*sin(27))/3.
Taking the inverse sine:
Q = sin^(-1)((5*sin(27))/3).
Q ≈ 23.67 degrees.
Therefore, there is 1 non-congruent triangle PQR that fits the given description, and angle Q is approximately 23.67 degrees.
To solve this problem, we can apply the Law of Cosines to find the length of side r, and then use the Law of Sines to find the non-congruent triangles PQR.
The Law of Cosines states that in any triangle with sides of lengths a, b, and c, and opposite angles A, B, and C respectively, the following equation holds:
c^2 = a^2 + b^2 - 2ab * cos(C)
Let's apply this to the given information.
(a) For triangle PQR, we have:
p = 3, q = 5, angle P = 27 degrees
Using the Law of Cosines, we can find the length of side r:
r^2 = p^2 + q^2 - 2pq * cos(P)
r^2 = 3^2 + 5^2 - 2*3*5 * cos(27°)
r ≈ 6.984
To find the non-congruent triangles, we will now use the Law of Sines, which states that in a triangle with sides of lengths a, b, and c, and opposite angles A, B, and C respectively, the following equation holds:
sin(A) / a = sin(B) / b = sin(C) / c
Since we know side p and angle P, we can solve for angle Q using the Law of Sines:
sin(27°) / r = sin(Q) / p
sin(Q) ≈ p * sin(27°) / r
Q ≈ arcsin(p * sin(27°) / r)
Substituting the values:
Q ≈ arcsin(3 * sin(27°) / 6.984)
Q ≈ 36.7°
Therefore, for case (a), there is only one non-congruent triangle and Q ≈ 36.7 degrees.
Now let's apply the same reasoning to the other cases:
(b) p = 8, q = 5, angle P = 57 degrees
Using the Law of Cosines, we find r ≈ 5.915
Using the Law of Sines, we find Q ≈ 33.8°
There is only one non-congruent triangle in this case.
(c) p = 7, q = 8, angle P = 70 degrees
Using the Law of Cosines, we find r ≈ 9.983
Using the Law of Sines, we find Q ≈ 20.0°
There is only one non-congruent triangle in this case.
(d) p = 10, q = 20, angle P = 30 degrees
Using the Law of Cosines, we find r ≈ 20.541
Using the Law of Sines, we find Q ≈ 44.5°
There is only one non-congruent triangle in this case.
I hope this explanation helps you understand how to find non-congruent triangles and the size of angle Q using the Law of Cosines and Law of Sines.
consider a right triangle PQR, with Q=90°. p = q sinP
So, in any other triangle, if given p,q, and P
if p < q sinP, then it is too short to meet side r, so no triangle possible
If p > q sinP, then there will be two triangles, one obtuse and one acute.
So, for problem a above, since 3 > 5 sin27°, there are two triangles that work
See what you can do with the others.