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. p=3,q=5,angleP=27degrees. p=8,q=5,angleP=57degrees. p=7,q=8,angleP=70degrees. p=10,q=20,angleP=30degrees
use the law of sines.
p/sinP = q/sinQ
if p = Q sinP then it is a right triangle
if p > Q sinP there are two triangles
if p < Q sinP there are no triangles
Oh boy, let's get ready for some triangular fun! Here we go:
1) p = 3, q = 5, angle P = 27 degrees.
Hmm, with sides 3 and 5, we know that it's definitely going to be a scalene triangle. Now, let's find out the number of non-congruent triangles and the size of angle Q.
Unfortunately, with the given information, we cannot determine the number of non-congruent triangles. However, we can use the Law of Sines to find angle Q:
sin Q / 5 = sin 27 / 3
Let's solve for sin Q:
sin Q = 5 * sin 27 / 3
sin Q ≈ 0.34301
Using the inverse sine function, we find:
Q ≈ 20.699 degrees
So, angle Q is approximately 20.699 degrees. Unfortunately, since we don't know the length of the third side, we can't construct a diagram.
2) p = 8, q = 5, angle P = 57 degrees.
Alrighty, here we go again! With sides 8 and 5, we know it's still going to be a scalene triangle. Yay for variety! Let's find the number of non-congruent triangles and the size of angle Q:
Since side p is longer than the sum of sides q and the base angle, there's only going to be one solution for this triangle.
Now, let's find angle Q using the Law of Cosines:
p^2 = q^2 + r^2 - 2qr * cos P
8^2 = 5^2 + r^2 - 2 * 5 * r * cos 57
64 = 25 + r^2 - 10r * cos 57
Substituting in the value of cos 57:
64 = 25 + r^2 - 10r * 0.55919
39 = r^2 - 5.5919r
Now, we need to solve this quadratic equation. After crunching the numbers:
r ≈ 1.641 or r ≈ 4.959
Since r must be greater than q, the only valid solution is r ≈ 4.959. Hence, there is only one non-congruent triangle in this case.
Now, let's find angle Q using the Law of Sines:
sin Q / 5 = sin 57 / 4.959
Let's solve for sin Q:
sin Q = 5 * sin 57 / 4.959
sin Q ≈ 0.82699
Using the inverse sine function, we find:
Q ≈ 56.475 degrees
So, angle Q is approximately 56.475 degrees. Now let's draw a diagram for this one:
P
/ \
8/ \5
/ \
Q-------R
3) p = 7, q = 8, angle P = 70 degrees.
This one's getting interesting! With sides 7 and 8, we know it's still going to be a scalene triangle. Let's find the number of non-congruent triangles and the size of angle Q:
Since side p is longer than the sum of sides q and the base angle, there's only going to be one solution for this triangle.
Now, let's find angle Q using the Law of Cosines:
p^2 = q^2 + r^2 - 2qr * cos P
7^2 = 8^2 + r^2 - 2 * 8 * r * cos 70
49 = 64 + r^2 - 16r * cos 70
Substituting in the value of cos 70:
49 = 64 + r^2 - 16r * 0.34202
-15 = r^2 - 5.4723r
Now, we need to solve this quadratic equation. After some number crunching:
r ≈ 3.388 or r ≈ 1.112
Since r must be greater than q, the only valid solution is r ≈ 3.388. Hence, there is only one non-congruent triangle in this case.
Now, let's find angle Q using the Law of Sines:
sin Q / 8 = sin 70 / 3.388
Let's solve for sin Q:
sin Q = 8 * sin 70 / 3.388
sin Q ≈ 1.04441
Uh-oh, something's fishy here! There's no solution for the equation above, so we can't determine the size of angle Q. But fear not, my friend, we can still draw a diagram:
P
/ \
7/ \8
/ \
Q-------R
4) p = 10, q = 20, angle P = 30 degrees.
Alright, last one! With sides 10 and 20, we know it's still going to be a scalene triangle. Let's find the number of non-congruent triangles and the size of angle Q:
Since side p is shorter than the sum of sides q and the base angle, there's going to be two non-congruent triangles in this case.
Now, let's find angle Q using the Law of Sines:
sin Q / 20 = sin 30 / 10
Let's solve for sin Q:
sin Q = 20 * sin 30 / 10
sin Q ≈ 0.34641
Using the inverse sine function, we find:
Q ≈ 20.877 degrees
So, angle Q is approximately 20.877 degrees. Now let's draw a diagram for this one:
P
/ \
10/ \20
/ \
Q-------R
Phew, that was quite the triangular journey! I hope my answers brought a smile to your face. If you have any more math questions or need a dose of humor, feel free to ask!
To solve each of these problems and find the number of non-congruent triangles PQR and the size of angle Q, we will use the Law of Cosines and the Law of Sines.
1. p = 3, q = 5, angle P = 27 degrees:
Using the Law of Cosines, we can find the length of the side r:
r^2 = p^2 + q^2 - 2pq * cos(P)
r^2 = 3^2 + 5^2 - 2 * 3 * 5 * cos(27)
r^2 = 9 + 25 - 30 * cos(27)
r^2 ≈ 35.2
r ≈ sqrt(35.2) ≈ 5.93
Since r is greater than the difference between p and q, there is only 1 non-congruent triangle that can be formed. Now, to find angle Q, we can use the Law of Sines:
sin(Q) / q = sin(P) / p
sin(Q) / 5 = sin(27) / 3
sin(Q) ≈ (5/3) * sin(27)
Q ≈ arcsin((5/3) * sin(27)) ≈ 46.2 degrees
The diagram for this case would show a triangle with sides of lengths 3, 5, and approximately 5.93, where angle P is 27 degrees and angle Q is approximately 46.2 degrees.
2. p = 8, q = 5, angle P = 57 degrees:
Using the Law of Cosines again:
r^2 = p^2 + q^2 - 2pq * cos(P)
r^2 = 8^2 + 5^2 - 2 * 8 * 5 * cos(57)
r^2 = 64 + 25 - 80 * cos(57)
r^2 ≈ 82.4
r ≈ sqrt(82.4) ≈ 9.08
Since r is greater than the difference between p and q, there is only 1 non-congruent triangle that can be formed. Now, using the Law of Sines:
sin(Q) / q = sin(P) / p
sin(Q) / 5 = sin(57) / 8
sin(Q) ≈ (5/8) * sin(57)
Q ≈ arcsin((5/8) * sin(57)) ≈ 29.0 degrees
The diagram for this case would show a triangle with sides of lengths 8, 5, and approximately 9.08, where angle P is 57 degrees and angle Q is approximately 29.0 degrees.
3. p = 7, q = 8, angle P = 70 degrees:
Using the Law of Cosines:
r^2 = p^2 + q^2 - 2pq * cos(P)
r^2 = 7^2 + 8^2 - 2 * 7 * 8 * cos(70)
r^2 = 49 + 64 - 112 * cos(70)
r^2 ≈ 95.4
r ≈ sqrt(95.4) ≈ 9.77
Since r is less than the sum but greater than the difference between p and q, there are 2 non-congruent triangles that can be formed. Now, using the Law of Sines:
sin(Q) / q = sin(P) / p
sin(Q) / 8 = sin(70) / 7
sin(Q) ≈ (8/7) * sin(70)
Q ≈ arcsin((8/7) * sin(70)) ≈ 53.5 degrees
The diagram for this case would show two triangles: one with sides of lengths 7, 8, and approximately 9.77, where angle P is 70 degrees and angle Q is approximately 53.5 degrees; and another triangle with sides of lengths 8, 7, and approximately 9.77, where angle P is 70 degrees and angle Q is also approximately 53.5 degrees.
4. p = 10, q = 20, angle P = 30 degrees:
Using the Law of Cosines:
r^2 = p^2 + q^2 - 2pq * cos(P)
r^2 = 10^2 + 20^2 - 2 * 10 * 20 * cos(30)
r^2 = 100 + 400 - 400 * cos(30)
r^2 ≈ 500
r ≈ sqrt(500) ≈ 22.36
Since r is greater than the sum of p and q, no triangle can be formed in this case.
Therefore, for p = 3, q = 5, angle P = 27 degrees, there is 1 non-congruent triangle and angle Q is approximately 46.2 degrees.
For p = 8, q = 5, angle P = 57 degrees, there is 1 non-congruent triangle and angle Q is approximately 29.0 degrees.
For p = 7, q = 8, angle P = 70 degrees, there are 2 non-congruent triangles and angle Q is approximately 53.5 degrees.
For p = 10, q = 20, angle P = 30 degrees, no triangle can be formed.
To determine the number of non-congruent triangles that fit a given description and find the size of angle Q, we can use the Law of Cosines and the Law of Sines. The Law of Cosines states that in a triangle, the square of one side is equal to the sum of the squares of the other two sides, minus twice the product of those two sides, and the cosine of the included angle. The Law of Sines states that in a triangle, the ratio of the length of a side to the sine of its opposite angle is constant.
Let's solve each case step by step:
Case 1: p = 3, q = 5, angle P = 27 degrees
To find the size of angle Q, we can use the Law of Cosines:
cos Q = (p^2 + q^2 - 2pq cos P) / (2pq)
cos Q = (3^2 + 5^2 - 2(3)(5) cos 27) / (2(3)(5))
cos Q = (9 + 25 - 30 cos 27) / 30
Now, find the value of Q by taking the inverse cosine (also known as arccosine) of cos Q:
Q = arccos(cos Q)
Finally, draw a triangle with sides of length 3 and 5, and angle P measuring 27 degrees, using a protractor or a drawing software. Measure angle Q with a protractor to determine its size.
Case 2: p = 8, q = 5, angle P = 57 degrees
Using the same process as in Case 1:
cos Q = (p^2 + q^2 - 2pq cos P) / (2pq)
cos Q = (8^2 + 5^2 - 2(8)(5) cos 57) / (2(8)(5))
cos Q = (64 + 25 - 80 cos 57) / 80
Q = arccos(cos Q)
Draw a triangle with sides of length 8 and 5, and angle P measuring 57 degrees, and measure angle Q.
Case 3: p = 7, q = 8, angle P = 70 degrees
cos Q = (p^2 + q^2 - 2pq cos P) / (2pq)
cos Q = (7^2 + 8^2 - 2(7)(8) cos 70) / (2(7)(8))
cos Q = (49 + 64 - 112 cos 70) / 112
Q = arccos(cos Q)
Draw a triangle with sides of length 7 and 8, and angle P measuring 70 degrees, and measure angle Q.
Case 4: p = 10, q = 20, angle P = 30 degrees
cos Q = (p^2 + q^2 - 2pq cos P) / (2pq)
cos Q = (10^2 + 20^2 - 2(10)(20) cos 30) / (2(10)(20))
cos Q = (100 + 400 - 400 cos 30) / 400
Q = arccos(cos Q)
Draw a triangle with sides of length 10 and 20, and angle P measuring 30 degrees, and measure angle Q.
Remember to use a protractor or accurate measuring tools when constructing and measuring the angles in the diagrams.