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In a triangle ABC, the lengths of the sides are a, b and c - Leaving Cert Mathematics - Question 5 - 2013
Question 5
In a triangle ABC, the lengths of the sides are a, b and c. Using a formula for the area of a triangle, or otherwise, prove that \[ \frac{a}{\sin A} = \frac{b}{\sin... show full transcript
Worked Solution & Example Answer:In a triangle ABC, the lengths of the sides are a, b and c - Leaving Cert Mathematics - Question 5 - 2013
Step 1
Prove that \( \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} \)
Answer
Using the formula for the area of a triangle, the area ( A ) can be expressed as:
[ A = \frac{1}{2} a b \sin C ]
Thus, we can equate the areas:
[ \frac{1}{2} a b \sin C = \frac{1}{2} \times c \times b \times \sin A \rightarrow \sin A = \frac{a \sin C}{c} ]
Dividing through by ( \sin A ), we obtain:
[ \frac{a}{\sin A} = \frac{c}{\sin C} ]
Applying the same logic for sides ( b ) and angles ( A, B ), we can also derive that ( \frac{b}{\sin B} ) holds, thus proving:
[ \frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} ]
Step 2
(i) Find the two possible values of |\angle XZY|
Answer
Using the Law of Sines:
[ \frac{XZ}{\sin Z} = \frac{XY}{\sin Y} ]
where |XZ| = 3 and |XY| = 5 cm, with |\angle XYZ| = 27°:
[ \frac{3}{\sin Z} = \frac{5}{\sin 27°} ]
Solving for ( \sin Z ):
[ \sin Z = \frac{3 \sin 27°}{5} \approx 0.756 ]
Thus, the two possible values for |\angle Z| are:
[ |Z| = 49° \text{ or } |Z| = 131° ]
Step 3
Step 4
In the case that |\angle XZY| < 90°, write down |\angle ZXY|
Answer
Using: [ |\angle ZXY| = 180° - (27° + 49°) = 104° ]
Now, to find the area of triangle XYZ:
[ A = \frac{1}{2} \times |XY| \times |XZ| \times \sin |\angle ZXY| ]
Substituting values: [ A = \frac{1}{2} \times 5 \times 3 \times \sin 104° \approx 7.27 ]
Thus, rounding to the nearest integer, the area of triangle XYZ is 7 cm².