How to construct triangle LMN , such that LN = 8 cm and ∠ LMN = 80 degrees and LM- MN= 3 cm?

Question

How to construct triangle LMN , such that LN = 8 cm and ∠ LMN = 80 degrees and LM- MN= 3 cm?

1 Answer

Impact of this question

3892 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2016-02-13T00:57:51

Use $L M = 7.461 c m$ $LM=7.461cm$ and $M N = 4.461 c m$ $MN=4.461cm$

Explanation:

Calling LM as $x$ $x$ and MN as $y$ $y$
$x - y = 3$ $x-y=3$ => $y = x - 3$ $y=x-3$

As $∠ L M N$ $/_ LMN$ is the angle between x and y
Applying the Law of Cosines:

$8^{2} = x^{2} + y^{2} - 2 x \cdot y \cdot {cos 80}^{\circ}$ $8^2=x^2+y^2-2x*y*cos 80^@$
$64 = x^{2} + {(x - 3)}^{2} - 2 x (x - 3) {cos 80}^{\circ}$ $64=x^2+(x-3)^2-2x(x-3)cos 80^@$
$x^{2} + x^{2} - 6 x + 9 - 2 x^{2} {cos 80}^{\circ} + 6 x {cos 80}^{\circ} - 64 = 0$ $x^2+x^2-6x+9-2x^2cos80^@+6xcos 80^@-64=0$
$(2 - 2 {cos 80}^{\circ}) x^{2} + (- 6 + 6 {cos 80}^{\circ}) x - 55 = 0$ $(2-2cos 80^@)x^2+(-6+6cos 80^@)x-55=0$

$Delta=36+36cos^2 80^@-72cos 80^@+440-440cos80^@$
$Delta=476+36cos^2 80^@-512cos80^@$
$Delta~=388.178$ => $sqrt(Delta)~=19.702$

$x=(6(1-cos 80^@)+-19.702)/(4(1-cos 80^@)$

$-> x_1=7.461cm$ => $y=7.461-3$ => $y=4.461cm$
$-> x_2=-4.461$ (not valid)

Science

Math

Humanities

... and beyond

How to construct triangle LMN , such that LN = 8 cm and ∠ LMN = 80 degrees and LM- MN= 3 cm?

1 Answer

Explanation:

Related questions

Impact of this question