# Find the height of a trapezoid

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 $$\measuredangle B.$$

(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!

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

First Solution

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

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

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

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

$$sin\angle{A}=\frac{5}{\sqrt{61}}$$

Second Solution

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

$$h^2=61-x^2=74-(13-x)^2$$

$$x=6, h=5$$

Third Solution

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

$$CP=AD=\sqrt{61}$$

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

$$h^2=61-x^2=74-(13-x)^2$$

$$x=6, h=5$$

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