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The diagram below shows part of the roof of a smaller shed (not to scale) - Leaving Cert Mathematics - Question e - 2022
Question e
The diagram below shows part of the roof of a smaller shed (not to scale). Some measurements are marked on the diagram. (i) Show that |BC| = 4.65 m, correct to 2 de... show full transcript
Worked Solution & Example Answer:The diagram below shows part of the roof of a smaller shed (not to scale) - Leaving Cert Mathematics - Question e - 2022
Step 1
Show that |BC| = 4.65 m, correct to 2 decimal places.
Answer
To find |BC|, we can use the cosine rule, given by:
[ c^2 = a^2 + b^2 - 2ab \cos(C) ]
In our case:
- Let |BC| be ( c ),
- With ( a = 3 ) m (side AC),
- ( b = 7 ) m (side AB),
- ( C = 30^{\circ} ).
Substituting these values:
[ c^2 = 3^2 + 7^2 - 2(3)(7) \cos(30^{\circ}) ] [ c^2 = 9 + 49 - 42 \times \frac{\sqrt{3}}{2} ] [ c^2 = 58 - 21.0 \sqrt{3} ] [ c^2 \approx 58 - 36.4411 \approx 21.5589 ]
Thus, [ c \approx \sqrt{21.5589} \approx 4.65 \text{ m (to 2 decimal places)} ]
Step 2
Find \angle ACB, the angle that the roof makes at the point C.
Answer
To find ( \angle ACB ), we can use the sine rule:
[ \frac{a}{\sin(A)} = \frac{b}{\sin(B)} = \frac{c}{\sin(C)} ]
Substituting the values we already have:
- Let |BC| = 4.65 m,
- Use ( a = 7 ),
- Find ( A = 30^{\circ} ).
So, [ \frac{7}{\sin(30^{\circ})} = \frac{4.65}{\sin(\angle ACB)} ] Since ( \sin(30^{\circ}) = 0.5 ):
[ \frac{7}{0.5} = \frac{4.65}{\sin(\angle ACB)} ] [ 14 = \frac{4.65}{\sin(\angle ACB)} ] [ \sin(\angle ACB) = \frac{4.65}{14} \approx 0.3321 ]
Thus, [ \angle ACB = \sin^{-1}(0.3321) \approx 19^{\circ} \text{ (to the nearest degree)} ]