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Here are two right-angled triangles - Edexcel - GCSE Maths - Question 20 - 2018 - Paper 3
Question 20
Here are two right-angled triangles. \[ 4x - 1 \] Given that \( \tan e = \tan f \) find the value of \( x \). You must show all your working.
Worked Solution & Example Answer:Here are two right-angled triangles - Edexcel - GCSE Maths - Question 20 - 2018 - Paper 3
Step 1
Given that \( \tan e = \tan f \)
Answer
To find the value of ( x ), we start by establishing the expression for ( \tan e ) and ( \tan f ) using the relationships in the right-angled triangles.
For triangle e: [ \tan e = \frac{\text{opposite}}{\text{adjacent}} = \frac{x}{4x - 1} ]
For triangle f: [ \tan f = \frac{\text{opposite}}{\text{adjacent}} = \frac{6x + 5}{12x + 31} ]
Setting ( \tan e ) equal to ( \tan f ): [ \frac{x}{4x - 1} = \frac{6x + 5}{12x + 31} ]
Step 2
Step 3
Step 4
Solving the quadratic equation
Answer
To solve the quadratic equation ( 12x^2 - 17x - 5 = 0 ), we can use the quadratic formula: [ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} ]
Where ( a = 12 ), ( b = -17 ), ( c = -5 ). Substituting these values gives: [ x = \frac{17 \pm \sqrt{(-17)^2 - 4 \cdot 12 \cdot (-5)}}{2 \cdot 12} ]
Calculating the discriminant: [ (-17)^2 - 4 \cdot 12 \cdot (-5) = 289 + 240 = 529 ]
Now substituting back: [ x = \frac{17 \pm \sqrt{529}}{24} = \frac{17 \pm 23}{24} ]
This gives two potential solutions: [ x = \frac{40}{24} = \frac{5}{3} \quad \text{or} \quad x = \frac{-6}{24} = -\frac{1}{4}]
Since ( x ) must be positive, we take ( x = \frac{5}{3} ).