5. (a) In the triangle ABC,
|AB| = 6 cm , |BC| = 5 cm
and |∠ABC| = 135° - Leaving Cert Mathematics - Question 5 - 2010
Question 5
5. (a) In the triangle ABC,
|AB| = 6 cm , |BC| = 5 cm
and |∠ABC| = 135°.
Calculate the area of the triangle,
correct to the nearest square centimetre.
(b) Cons... show full transcript
Worked Solution & Example Answer:5. (a) In the triangle ABC,
|AB| = 6 cm , |BC| = 5 cm
and |∠ABC| = 135° - Leaving Cert Mathematics - Question 5 - 2010
Step 1
(a) Calculate the area of the triangle
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Answer
To find the area of triangle ABC, we can use the formula:
Area=21∣AB∣∣BC∣sin∠ABC
Substituting in the known values:
Area=21×6×5×sin(135°)
Since ( \sin(135°) = \sin(45°) = \frac{\sqrt{2}}{2} ), we continue:
Area=21×6×5×22
Calculating this gives:
Area=152≈21.21extcm2
Rounding this to the nearest square centimetre, we find:
Area≈21extcm2
Step 2
(b) (i) Find the value of x.
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Answer
Using the Pythagorean theorem:
x2+152=172
Calculating gives:
x2+225=289⇒x2=64⇒x=8
Step 3
(b) (ii) Write down, as a fraction, the value of sin θ.
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Answer
The sine of an angle in a right triangle is given by the opposite side over the hypotenuse:
sin(θ)=1715
Step 4
(b) (iii) Write down, as a fraction, the value of cos θ.
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Answer
The cosine of an angle is given by the adjacent side over the hypotenuse:
cos(θ)=178
Step 5
(b) (iv) Find the value of sin 2θ.
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Answer
Using the double angle formula:
sin(2θ)=2sin(θ)cos(θ)
Substituting the known values:
sin(2θ)=2×1715×178=289240
Step 6
(c) (i) Find |PR|, correct to the nearest metre.
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Answer
Using the cosine rule:
∣PR∣2=∣PQ∣2+∣QR∣2−2∣PQ∣∣QR∣cos(50°)
The unknown length is:
∣PR∣=17.42+∣PQ∣2−2×17.4×∣PQ∣cos(50°)
Since |PQ| is still unknown, we cannot compute further without additional information about PQ.
Step 7
(c) (ii) Find |PS|, correct to the nearest metre.
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Answer
Using the Pythagorean theorem:
∣PS∣=∣PQ∣2+∣SR∣2
Substituting the known values:
∣PS∣=∣PQ∣2+152=∣PQ∣2+225
Similarly, without information on |PQ|, we cannot compute further.
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