1. **Problem statement:** (i) Calculate the length of |AB| in the triangular truss where |AC|=6 m and the roof pitch angle at A is 35°. (ii) Calculate the total length of timber required for the truss. (iii) In triangle PQR with sides |PQ|=91 cm, |QR|=125 cm, and |PR|=204 cm, find the angle |\angle PQR|. 2. **Step (i): Calculate |AB|** - Given |AC|=6 m and pitch angle $\angle BAC=35^\circ$. - Since |AD|=|DE|=|EC|, each segment is $\frac{6}{3}=2$ m. - Point B is above the base AC, and the vertical from B to DE is perpendicular. - Triangle ABD is right-angled at D, with |AD|=2 m and $\angle BAD=35^\circ$. - Use trigonometry: $\tan(35^\circ) = \frac{|BD|}{|AD|} \Rightarrow |BD| = |AD| \times \tan(35^\circ) = 2 \times \tan(35^\circ)$. - Calculate $\tan(35^\circ) \approx 0.7002$, so $|BD| \approx 2 \times 0.7002 = 1.4004$ m. - Now, |AB| is the hypotenuse of right triangle ABD: $$|AB| = \sqrt{|AD|^2 + |BD|^2} = \sqrt{2^2 + 1.4004^2} = \sqrt{4 + 1.9611} = \sqrt{5.9611} \approx 2.44\text{ m}.$$ 3. **Step (ii): Calculate total timber length** - Given |AF|=|FB|=|BG|=|GC|, so the base AB is divided into 4 equal parts. - Since |AB| $\approx 2.44$ m, each segment is $\frac{2.44}{4} = 0.61$ m. - The truss consists of segments: AC (6 m), AB (2.44 m), BC (equal to AB by symmetry, so 2.44 m), plus internal segments AF, FB, BG, GC (4 segments each 0.61 m), and AD, DE, EC (3 segments each 2 m). - Total length: $$6 + 2.44 + 2.44 + 4 \times 0.61 + 3 \times 2 = 6 + 2.44 + 2.44 + 2.44 + 6 = 19.32\text{ m}.$$ 4. **Step (iii): Find $\angle PQR$ in triangle PQR** - Use the Law of Cosines: $$|PR|^2 = |PQ|^2 + |QR|^2 - 2|PQ||QR|\cos(\angle PQR)$$ - Substitute values: $$204^2 = 91^2 + 125^2 - 2 \times 91 \times 125 \times \cos(\angle PQR)$$ - Calculate squares: $$41616 = 8281 + 15625 - 22750 \cos(\angle PQR)$$ - Simplify: $$41616 = 23906 - 22750 \cos(\angle PQR)$$ - Rearrange: $$22750 \cos(\angle PQR) = 23906 - 41616 = -17710$$ - So: $$\cos(\angle PQR) = \frac{-17710}{22750} \approx -0.7787$$ - Find angle: $$\angle PQR = \cos^{-1}(-0.7787) \approx 141.0^\circ.$$