Given a trapezoid like the one shown below, how do I determine the length of the wider base? I'm looking for a formula based method rather than drawing the shape and measuring it.
[Math] How to determine the length of the wider base of a trapezoid from the height + shorter base length + angles
geometrytrigonometry
Related Solutions
[Math] How to compute the length of line segments inside an isosceles trapezoid parallel to the base
Suppose the top line is $ \overline{AB}$ and the bottom line is $\overline{CD}$. Now, find a point $E$ in $\overline{CD}$ such that $\overline{AE}$ is parallel to $\overline{BD}$. Then apply the rule for the ratio of similar triangles to the triangle $\overline{ACE}$.
Update: $$\frac{AB}{AD}=\frac{AC}{AE}=\frac{BC}{DE}$$ if $\triangle ABC$ and $\triangle ADE$ are similar triangles.
Update 2: I forget to tell you another fact. If $\triangle ABC$ and $\triangle ADE$ are similar triangles, $$\frac{\text{height of }\triangle ABC}{\text{height of } \triangle ADE}=\frac{AB}{AD}$$
$$a = b - 2 \,c \cos \alpha $$
where $ \alpha = $ angle between the big side and one of equal sides length $c$.
Related Question
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Best Answer
Use the Law of Sines on the triangles (separately) on either end.
For example, on the right side of the trapezoid, drop a perpendicular from the top right vertex. Call the length of the base of the resulting right triangle $x$. The other (non-hypotenuse) side is 3/4. The bottom (interior) angle is $70^\circ$ while the top (interior) angle is $20^\circ$.
The Law of Sines says $${{3\over 4}\over \sin(70^\circ)}={x\over \sin(20^\circ)}.$$ Solve for $x$ to obtain $x={3\over 4}\cdot{\sin(20^\circ)\over \sin(70^\circ)}\approx 0.272978$.
Play the same game on the other side and you will know the total length of the base of the trapezoid.