Question

1. a scissor truss is shown in figure 23 - 7. the sizes of \\( \angle c a b = 16 ^ { \circ } \\) and \\( \angle e c d = 34 ^ { \circ } \\).

determine each of the following angles.
a. \\( \angle a b c \\) b. \\( \angle a c b \\) c. \\( \angle c b d \\) d. \\( \angle b d c \\) e. \\( \angle c e d \\)

Explanation:

Step1: Find $\angle ABC$

Since the angle at the joint of the truss is a right - angle (implicit from the figure structure), in triangle \(ABC\), \(\angle ABC = 90^{\circ}\)

Step2: Find $\angle ACB$

In \(\triangle ABC\), using the angle - sum property of a triangle (\(\angle A+\angle B+\angle C = 180^{\circ}\)). Given \(\angle A=16^{\circ}\) and \(\angle B = 90^{\circ}\).

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Step3: Find $\angle CBD$

\(\angle CBD\) and \(\angle ABC\) are supplementary (linear pair). So \(\angle CBD=180^{\circ}-\angle ABC\)
Since \(\angle ABC = 90^{\circ}\), \(\angle CBD = 90^{\circ}\)

Step4: Find $\angle BDC$

In \(\triangle BCD\), \(\angle BCD = 180^{\circ}-\angle ACB\) (linear pair). \(\angle BCD=180 - 74=106^{\circ}\), \(\angle CBD = 90^{\circ}\)
Using \(\angle BDC+\angle CBD+\angle BCD=180^{\circ}\)

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Wait, actually, since \(BC\perp CD\) (from the truss structure, right - angle at \(C\) for the lower part related to \(BD\)), \(\angle BCD = 90^{\circ}\) (corrected based on truss right - angle property). Then \(\angle BDC=180^{\circ}-\angle CBD-\angle BCD\) (where \(\angle CBD = 90^{\circ}\), \(\angle BCD = 90^{\circ}-\angle ECD\), \(\angle ECD = 34^{\circ}\), so \(\angle BCD=56^{\circ}\))

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Step5: Find $\angle CED$

In \(\triangle CED\), \(\angle ECD = 34^{\circ}\), \(\angle EDC = 90^{\circ}\) (right - angle from truss structure)
Using \(\angle CED+\angle ECD+\angle EDC=180^{\circ}\)

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Answer:

a. \(90^{\circ}\)
b. \(74^{\circ}\)
c. \(90^{\circ}\)
d. \(34^{\circ}\)
e. \(56^{\circ}\)