I presume that \(x\) refers to the angle between the two intersecting lines.

Firstly, notice that for every two intersecting lines in a Cartesian plane, there will be one line which makes a larger angle with the positive x-axis.

We shall call the gradient of this line as \(m_1\). Similarly, we shall call the gradient of the line which makes a smaller angle with the positive x-axis as \(m_2\)

The relation between the angle and the gradient is given by

\[\tan x=\frac{m_1-m_2}{1+m_1m_2}\]
–
Ho Wei Haw
·
3 years, 1 month ago

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@Ho Wei Haw
–
i did not understand the Penultimate equation
–
Abdou Ali
·
3 years, 1 month ago

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@Abdou Ali
–
It is the different of angles formula for tan. For example, \(\tan(a-b)=\dfrac{\tan a - \tan b}{1+\tan a \tan b}\).
–
Daniel Liu
·
3 years, 1 month ago

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TopNewestI presume that \(x\) refers to the angle between the two intersecting lines.

Firstly, notice that for every two intersecting lines in a Cartesian plane, there will be one line which makes a larger angle with the positive x-axis.

We shall call the gradient of this line as \(m_1\). Similarly, we shall call the gradient of the line which makes a smaller angle with the positive x-axis as \(m_2\)

The relation between the angle and the gradient is given by

\[m_1 = \tan \theta_1\]

\[m_2 = \tan \theta_2\]

where \(\theta_1>\theta_2\)

Notice that \(x=\theta_1-\theta_2\)

So, taking tangents on both sides,

\[\tan x=\tan(\theta_1-\theta_2)\]

By the addition formula for tangent,

\[\tan x=\frac{\tan\theta_1-\tan\theta_2}{1+\tan\theta_1\tan\theta_2}\]

which is equivalent to

\[\tan x=\frac{m_1-m_2}{1+m_1m_2}\] – Ho Wei Haw · 3 years, 1 month ago

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– Abdou Ali · 3 years, 1 month ago

i did not understand the Penultimate equationLog in to reply

– Daniel Liu · 3 years, 1 month ago

It is the different of angles formula for tan. For example, \(\tan(a-b)=\dfrac{\tan a - \tan b}{1+\tan a \tan b}\).Log in to reply