Easy Geometry

Geometry Level 4

ABCD is a square on a side of 0.96 metre , and from Q its geometric centre a rod QP of length 1.40 metre is fixed perpendicular to the plane. If R is the mid-point of the side BC, calculate the value of cos ⁡ P R Q \cos \ PRQ correct to the third decimal place .

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The answer is 0.324.

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4 solutions

Shariful Islam
Jan 2, 2015

QR=AB/2=0.96/2 = 0.48 metre and PR= SQRT (0.48^2+1.4^2) =1.48 metre So cosPRQ=QR/PR=0.48/1.48 = 0.324324

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0.324324....=12/37

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Vishnu Bhagyanath
Jul 12, 2015

Given Q R = 0.48 QR = 0.48 , Q P = 1.4 QP=1.4

To find, c o s ( t a n − 1 P R Q ) cos (tan^{-1} PRQ) = c o s ( t a n − 1 ( 1.4 / 0.48 ) = cos ( tan^{-1} (1.4/0.48 ) ≈ 0.324 \approx 0.324

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Anmol Parande
Jan 1, 2015

The length of RQ is 0.48 because the distance from a midpoint of a square's side to the center is half the length of the side. The problem states PQ is 1.4. It also says that PQ is perpendicular to the square, therefore we can use pythagorean theorem to find PR. 0.48^2 + 1.4^2 = 2.1094. The square root of 2.1904 = 1.48 therefore RP is 1.48. Cosine is opposite/adjacent so cosPRQ is frac 0.48/1.48 which is 0.324 when rounded to the thousandths place

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