angle wanted in a cyclic ABCD with sidelenghts 6,3,5,4

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angle wanted in a cyclic ABCD with sidelenghts 6,3,5,4

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Source: All-Siberian Open School Olympiad 2013-14 11.1

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#1 h Jul 2, 2020, 5:51 PM

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The lengths of the sides of the inscribed quadrangle in the clockwise order are $6,3, 5,4$ , respectively. Find the angle between sides of length $6$ and $3$ .

This post has been edited 1 time. Last edited by parmenides51, Jul 2, 2020, 5:58 PM

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Kuttus

#2 h Jul 2, 2020, 6:33 PM

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name the lengths of the vertex as A,B,C and D such that AB=6 and then in order. Draw the Diagonal AC and apply cosine rule in ABC and ACD to get a common expression for AC^2.Equate them to get the value of the angle.

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Neo-Pythagorean

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Neo-Pythagorean

#3 h Jul 2, 2020, 7:13 PM

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Let $ABCD$ be the inscribed quadrilateral of the problem with $AB=6$ . From Ptolemy's Theorems we get:
$AC*BD=AB*CD+AD*BC=6*5+3*4=42$ $\frac{AC}{BD}=\frac{AB*AD+CB*CD}{BA*BC+DA*DC}=\frac{6*4+3*5}{6*3+4*5}=\frac{39}{38}$ Hence, $AC^2=\frac{819}{19}$ and from the law of cosines in $\triangle{ABC}$ , we have:
$AC^2=AB^2+BC^2-2AB*BC*cos(\angle{B}) \Longleftrightarrow \frac{819}{19}=36+9-36*cos(\angle{B}) \Longleftrightarrow$ $cos(\angle{B})=\frac{1}{19}$

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