Points $E,B,D$ are collinear. Given $BD=10$ cm, $BC=12$ cm, \(\angle CEB=\angle ABD=90^\c

Question

Points $E,B,D$ are collinear. Given $BD=10$ cm, $BC=12$ cm, $\angle CEB=\angle ABD=90^\circ$, $\angle ADB=40^\circ$ and $\angle CBD=130^\circ$, calculate $AB$, $CE$, the area of $\triangle BCD$. You may use $\sin40^\circ=0.643$, $\cos40^\circ=0.766$, $\tan40^\circ=0.839$, $\sin50^\circ=0.766$, $\cos50^\circ=0.643$. Give non-exact lengths, areas and other numerical values to 3 significant figures, and angles to 1 decimal place, unless otherwise stated.

Accepted Answer

(a) $8.39$ cm; (b) $9.19$ cm; (c) $46.0\text{ cm}^2$.

Two right triangles around a straight line

Two right triangles around a straight line

Solution:

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