Asked by Insyirah
In the figure, QRST isa a square and PQT is an equilateral triangle. Find PRT? So it looks like a house and TSR is 90 degrees.
Answers
Answered by
oobleck
If we let RS = 2, then
the altitude of ΔPTQ = √3
and so PR^2 = (2+√3)^2 + 1^2 = 8+4√3
TR = 2√2
PT = 2
That means that using the law of cosines,
2^2 = 8+4√3 + 8 - 2√(8+4√3)(√8) cos(PRT)
cos(PRT) = 0.86603
so PRT is very nearly 30°
to verify this, do the construction and measure the angle.
the altitude of ΔPTQ = √3
and so PR^2 = (2+√3)^2 + 1^2 = 8+4√3
TR = 2√2
PT = 2
That means that using the law of cosines,
2^2 = 8+4√3 + 8 - 2√(8+4√3)(√8) cos(PRT)
cos(PRT) = 0.86603
so PRT is very nearly 30°
to verify this, do the construction and measure the angle.
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