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4 years ago

Construct a parallelogram ABCD such that AB=9 Cm, BC=6cm and ABC=115 degress.

Mathematics

2 answers:

lara31 [8.8K]4 years ago

6 0

Answer:

1.BD=AC

AC^2=(9^2)+(6^2)-2(9)(6)COS115

AC^2=117-108COS115

AC=√71.358

AC=8.447//1.BD=AC

AC^2=(9^2)+(6^2)-2(9)(6)COS115

AC^2=117-108COS115

AC=√71.358

AC=8.447//

Step-by-step explanation:

kkurt [141]4 years ago

3 0

1.BD=AC
AC^2=(9^2)+(6^2)-2(9)(6)COS115
AC^2=117-108COS115
AC=√71.358
AC=8.447//

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<h3>Solution:</h3>

Given that line m contains points (-3, 4) and (1, 0)

We are asked to find the equation of line perpendicular to line containing points (-3, 4) and (1, 0) and passes through (-2, 6)

Let us first find slope of the line "m"

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We know that product of slope of given line and perpendicular line are always -1

So, we get

$\begin{array}{l}{\text { slope of line } m \times \text { slope of perpendicular line }=-1} \\\\ {-1 \times \text { slope of perpendicular line }=-1} \\\\ {\text { slope of perpendicular line }=1}\end{array}$

So we have got the slope of perpendicular line is 1 and it passes through (-2, 6)

Let us use the point slope form to find the required equation

The point slope form is given as:

$y - y_1 = m(x - x_1)$

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