In triangle PQR, m<P=53 degrees, PQ=7.4, and PR=9.6. What is m<R to the nearest degree
To find the measure of angle R, you can use the fact that the sum of the angles in a triangle is always 180 degrees.
Let's denote the measure of angle R as m<R.
Since you know that angle P has a measure of 53 degrees and the sum of the angles in a triangle is 180 degrees, you can use the equation:
m<P + m<Q + m<R = 180
Substituting the given value:
53 + m<Q + m<R = 180
To solve for m<R, you need to find the measure of angle Q. Since you only have the lengths of two sides, you can use the Law of Cosines to find angle Q.
The Law of Cosines states that for a triangle with sides of lengths a, b, and c, and an opposite angle C, the equation is:
c^2 = a^2 + b^2 - 2ab * cos(C)
In this case, side QR has length 7.4 and side PR has length 9.6. The angle opposite side QR is angle P, and the angle opposite side PR is angle Q.
Using the Law of Cosines:
7.4^2 = 9.6^2 + 7.4^2 - 2 * 9.6 * 7.4 * cos(Q)
54.76 = 92.16 + 54.76 - 2 * 9.6 * 7.4 * cos(Q)
2 * 9.6 * 7.4 * cos(Q) = 92.16
cos(Q) = 92.16 / (2 * 9.6 * 7.4)
cos(Q) ≈ 0.707
Now, using the inverse cosine function (arccos) on a calculator, you can find that Q ≈ 45 degrees.
Substituting the value of Q into the equation for the sum of angles:
53 + 45 + m<R = 180
98 + m<R = 180
To isolate m<R, subtract 98 from both sides:
m<R = 180 - 98
m<R ≈ 82 degrees
Therefore, the measure of angle R is approximately 82 degrees.
To find the measure of angle R in triangle PQR, we can use the fact that the sum of the measures of angles in any triangle is always 180 degrees.
In this case, we know that angle P is 53 degrees. So, to find angle R, we need to subtract the measures of angles P and Q from 180 degrees.
Step 1: Find the measure of angle Q.
The sum of angles P and Q is 180 - 53 degrees, which gives us 127 degrees.
So, angle Q is 127 degrees.
Step 2: Find the measure of angle R.
To find the measure of angle R, subtract the sum of angles P and Q from 180 degrees:
180 - (53 + 127) = 180 - 180 = 0 degrees.
Since a triangle must have a non-zero measure for all its angles, it seems that we've made an error. It's not possible for angle R in triangle PQR to have a measure of 0 degrees.
Please double-check the given information and try again.
first what is QR which is p?
law of cosines
p^2 = q^2 + r^2 - 2 p q cos P
p^2 = 9.6^2 + 7.4^2 - 2(7.4)(9.6)cos 53
solve for p
then law of sines
sin R/7.4 = sin 53/ p