Find an equation for the plane containing the point (1, 2, 3) and parallel to the plane x 2y 3z
1 answer:
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Part A:
From the figure shown.
The measure of line TH = 14' 72" = 14' + 72" / 12 = 14' + 6' = 20'
Given that the measure of line HI = 20'
Thus, triangle THI is an isosceles triangle with line TI as the base.
Part B:
To find the base angles we use the Pythagoras theorem.
Recall that the perpendicuar bisector of an isosceles triangle divides the triangle into two equal right triangles.
Dividing the base into two gives 22.5' / 2 = 11.25'
Thus, the perpendicular bisector divides the isosceles triangle into two equal right triangles with base of 11.25' and hypothenus of 20'.
By pythagoras theorem,
Therefore, the base angles is 55.77 degrees.
From the figure shown.
The measure of line TH = 14' 72" = 14' + 72" / 12 = 14' + 6' = 20'
Given that the measure of line HI = 20'
Thus, triangle THI is an isosceles triangle with line TI as the base.
Part B:
To find the base angles we use the Pythagoras theorem.
Recall that the perpendicuar bisector of an isosceles triangle divides the triangle into two equal right triangles.
Dividing the base into two gives 22.5' / 2 = 11.25'
Thus, the perpendicular bisector divides the isosceles triangle into two equal right triangles with base of 11.25' and hypothenus of 20'.
By pythagoras theorem,
Therefore, the base angles is 55.77 degrees.