Eric Nicholas K's answer to SK's Secondary 4 A Maths Singapore question.
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Here is one possible approach for this question.
If you have not already known it, the gradient of a line is closely linked to the angle at which the line makes with the horizontal.
Since the gradient of the line is all about the height of the triangle divided by the base of the triangle, and the tangent of this angle is the height of the triangle (opposite) divided by the base of the triangle (adjacent), it follows that both the gradient and the tangent of the angle are more or less almost similar to each other.
There is another approach in a very similar manner to part i if you are not comfortable with this approach. This second approach requires you to find the equation of line PR given the gradient -13/8 and point P (8, -8) and solve the equations QR and PR simultaneously to find the coordinates of R, before finding any two lengths in triangle PQR and performing “Toa Cah Soh” to extract angle PRQ and thereafter angle theta.
If you have not already known it, the gradient of a line is closely linked to the angle at which the line makes with the horizontal.
Since the gradient of the line is all about the height of the triangle divided by the base of the triangle, and the tangent of this angle is the height of the triangle (opposite) divided by the base of the triangle (adjacent), it follows that both the gradient and the tangent of the angle are more or less almost similar to each other.
There is another approach in a very similar manner to part i if you are not comfortable with this approach. This second approach requires you to find the equation of line PR given the gradient -13/8 and point P (8, -8) and solve the equations QR and PR simultaneously to find the coordinates of R, before finding any two lengths in triangle PQR and performing “Toa Cah Soh” to extract angle PRQ and thereafter angle theta.
Date Posted:
4 years ago
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