A trapezoid $ABCD$ is given $(AB$ $||$ $CD)$ with side lengths $AB=18,BC=\sqrt{74},CD=5$ and $AD=\sqrt{61}.$ Find the sines of $\measuredangle A$ and $\measuredangle B.$

enter image description here (I have only studied trig functions of acute angles) Since we are given the four sides of the trapezoid, it is enough to find the height. I am not sure how to approach the problem. If $CP$ is parallel to $AD,$ we can find the area of triangle $PBC$ by Heron's formula (the area of a triangle when the length of all three sides are known) but we still haven't studied it. Can you give me a hint for another solution? Thank you in advance!

  • $\begingroup$ Have you tried finding $AD_1$ and $BC_1$? $\endgroup$ – Jose Ramirez Sep 24 '20 at 19:44
  • $\begingroup$ Hint : 36+25=61. 49+25=74. 6+5+7=18 $\endgroup$ – cosmo5 Sep 24 '20 at 19:48

First Solution

$PC=AD=\sqrt{61}$, $BC=\sqrt{74}$, $BP=13$

$cos \angle{B}=\frac{74+169-61}{26\sqrt{74}}=\frac{7}{\sqrt{74}}$

$sin \angle{B}=\frac{5}{\sqrt{74}}$

$cos \angle{A}= cos \angle{P}=\frac{61+169-74}{26\sqrt{61}}=\frac{6}{\sqrt{61}}$


Second Solution

Let $AD_1=x, BC_1=13-x, CC_1=DD_1=h$


$x=6, h=5$

Third Solution

$BP=13$ since $AP=CD=5$


Let $PC_1=x$, $BC_1=13-x$, $CC_1=h$


$x=6, h=5$

  • $\begingroup$ Thank you for the response! I have only studied trig functions of acute angles. $\endgroup$ – Katherine Sep 24 '20 at 19:43
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    $\begingroup$ I have sent alternative solution it will help you to find height of trapezium $\endgroup$ – Lion Heart Sep 24 '20 at 19:59
  • $\begingroup$ Why $BC_1=13-x$? If $AD_1$ is $x$, then $13-x$ is $BD_1$. $\endgroup$ – Katherine Sep 24 '20 at 20:07
  • $\begingroup$ $D_1C_1=DC=5$ and $CC_1D_1D$ is a rectangle since $DD_1\perp AB$, $CC_1\perp AB$ and $AB\parallel CD$ $\endgroup$ – Lion Heart Sep 24 '20 at 20:18
  • $\begingroup$ I have sent another one, using triangle CPB finding height of trapezium $\endgroup$ – Lion Heart Sep 24 '20 at 20:47

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