+150 In Triangle SPT, Q is on SP and R is on PT. We know PQ = 8, QS = 4, ST = 9, and PR = QR = 6. Find all possible values of RT.
There was no picture. Thank you if you could help.
+129852 We have something like this
P
8 6
R
Q
4
S 9 T
We know three sides of isisceles triangle PQR, PQ = 8 and PR and QR = 6
By the Law of Cosines we can find angle SPT= angle QPR
QR^2 = PQ^2 + PR^2 - 2(PQ)(PR) cos ( SPT)
6^2 = 8^2 + 6^2 - 2(8)(6) cos (SPT)
-8^2 = - 96 cos (SPT)
64/96 = cos (SPT)
SPT = arccos ( 64/96) ≈ 48.19“
In triangle SPT, we know an angle (SPT) and two sides SP and ST
We have SSA (an ambiguous case).....we may have no triangle......one triangle...or two triangles
Using the Law of Sines
sin (SPT) / ST = sin PTS / PS
sin (48.2) / 9 = sin PTS / 12
(4/3)sin 48.2 = sin PTS
arcsin ( 4/3 * sin 48.2) = PTS ≈ 83.7°
Then angle PST = 180 - 48.2 - 83.7 = 48.1°
And by the Law of Sines we can find PT as
PT/ sin PST = ST / sin SPT
PT / sin 48.1 = 9 / sin 48.2
PT = 9 sin 48.2/sin 48.1
PT ≈ 9
So RT = PT - PR = 9 - 6 = 3
We may also have another triangle
Since the sine of an angle = the sine of its supplement
supplement of PTS = (180 - 83.7) = 96.3°
So the remaining angle, PST could = 180 - 48.2 - 96.3 = 35.5°
So....usng the Law of Sines again
PT/ sin PST = ST / sin SPT
PT / sin 35.5 = 9/ sin 48.2
PT = 9 sin 35.5/sin48.2 ≈ 7
And RT could also take the value PT -- PR = 7 - 6 = 1
