Answer :
Answer:
(a) 128.9°
(b) 41.1 m
Step-by-step explanation:
Part (a)
To find the size of angle ACB, we can apply the Law of Sines to triangle ABC.
[tex]\boxed{\begin{array}{l}\underline{\textsf{Law of Sines}} \\\\\dfrac{\sin A}{a}=\dfrac{\sin B}{b}=\dfrac{\sin C}{c}\\\\\textsf{where:}\\\phantom{ww}\bullet \;\textsf{$A, B$ and $C$ are the angles.}\\\phantom{ww}\bullet\;\textsf{$a, b$ and $c$ are the sides opposite the angles.}\end{array}}[/tex]
In triangle ABC:
- ∠CAB = 27°
- AB = 12 m
- BC = 7 m
As angle ACB is obtuse:
[tex]\dfrac{\sin CAB}{BC}=\dfrac{\sin(180^{\circ}-ACB)}{AB}[/tex]
Substitute the given values into the equation and solve for angle ACB:
[tex]\dfrac{\sin 27^{\circ}}{7}=\dfrac{\sin(180^{\circ}-ACB)}{12} \\\\\\ \sin(180^{\circ}-ACB)=\dfrac{12\sin 27^{\circ}}{7} \\\\\\ 180^{\circ}-ACB=\sin^{-1}\left(\dfrac{12\sin 27^{\circ}}{7}\right) \\\\\\ ACB=180^{\circ}-\sin^{-1}\left(\dfrac{12\sin 27^{\circ}}{7}\right) \\\\\\ ACB=128.897602169211...^{\circ} \\\\\\ACB=128.9^{\circ} \; \sf (1\;d.p.)[/tex]
Therefore, the size of angle ACB rounded to one decimal place is:
[tex]\LARGE\boxed{\boxed{ACB=128.9^{\circ}}}[/tex]
[tex]\dotfill[/tex]
Part (b)
The minimum length of steel that needs to be bought to make the complete structure is the sum of lengths AB, BD, BC and AD.
We are given that AB = 12 m and BC = BD = 7 m, so we need to find the length of AD. To do that, we first need to find the measure of angle ABD of triangle ABD.
Angles ACB and BCD form a linear pair, so:
[tex]\angle ACB + \angle BCD=180^{\circ}\\\\ \angle BCD=180^{\circ}-\angle ACB[/tex]
Since BC = BD, triangle BCD is an isosceles triangle. The angles opposite the congruent sides in an isosceles triangle are equal, so ∠BCD = ∠CDB. Therefore:
[tex]\angle CDB=180^{\circ}-\angle ACB[/tex]
Since ∠CDB = ∠ADB, then:
[tex]\angle ADB=180^{\circ}-\angle ACB[/tex]
Angles in a triangle sum to 180°, so in triangle ABD:
[tex]\angle ADB+\angle BAD+\angle ABD=180^{\circ} \\\\180^{\circ}-\angle ACB+27^{\circ}+\angle ABD=180^{\circ} \\\\ \angle ABD=\angle ACB-27^{\circ} \\\\ \angle ABD=128.897602169211...^{\circ}-27^{\circ} \\\\ \angle ABD=101.897602169211...^{\circ}[/tex]
Now, we can use the Law of Sines again to find the length of AD:
[tex]\dfrac{\sin ABD}{AD}=\dfrac{\sin BAD}{BD} \\\\\\ \dfrac{\sin 101.897602169211...^{\circ}}{AD}=\dfrac{\sin 27^{\circ}}{7} \\\\\\ AD=\dfrac{7\sin 101.897602169211...^{\circ}}{\sin 27^{\circ}} \\\\\\ AD=15.087591703396... \\\\\\ AD=15.1\; \sf m\;(1\;d.p.)[/tex]
Finally, add together AB, BD, CD and AD:
[tex]\textsf{Length of steel}=AB + BD +BC + AD\\\\\textsf{Length of steel}=12+7+7+15.1 \\\\\textsf{Length of steel}=41.1\; \sf m\;(1\;d.p.)[/tex]
Therefore, the minimum length of steel that needs to be bought to make the complete structure is:
[tex]\LARGE\boxed{\boxed{41.1\; \sf m}}[/tex]
