Mathematics: In triangle IJK, J = 800cm, k = 240 cm and angle I=173 degrees. Find the area of triangle IJK, to the nearest square centimeter.

In triangle IJK, J = 800cm, k = 240 cm and angle - id80024012085445, sugandiwoodbine

Ответ:

journeyburks07

19.04.2021

Mathematics

11,700cm²

Step-by-step explanation:

Area of the triangle = 1/2*JK sin<I

Given

J = 800cm

K = 240cm

<I = 173 degrees

Substitute

Area of the triangle = 1/2 * 800 * 240 sin 173

Area of the triangle = 400 * 240sin 173

Area of the triangle = 400 * 29.25

Area of the triangle = 11,699.5

Area of the triangle to the nearest square centimeter is 11,700cm²

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Ответ:

itzzarosario12

17.05.2022

Mathematics

$\rm x=\frac{46}{3}$

$\rm y=-\frac{25}{24}$

Step-by-step explanation:

$\huge\begin{array} {|c|} \hline\left. \begin{cases} { \rm\frac{ 1 }{ 4 } x+ \frac{ 4 }{ 5 } y = 3 } \\ { \rm\frac{ 5 }{ 16 } x- \frac{ 1 }{ 5 } y=5 } \end{cases} \right. \\ \hline \end{array}$

$\\ \Large \overline{ \rm - \: Steps \: using \: substitution \: - }$

$\rm \frac{1}{4}x+\frac{4}{5}y=3$

$\rm \frac{1}{4}x=-\frac{4}{5}y+3$

$\rm x=4\left(-\frac{4}{5}y+3\right)$

$\rm x=-\frac{16}{5}y+12$

┈

$\rm \frac{5}{16}\left(-\frac{16}{5}y+12\right)-\frac{1}{5}y=5$

$\rm -y+\frac{15}{4}-\frac{1}{5}y=5$

$\rm -\frac{6}{5}y+\frac{15}{4}=5$

$\rm -\frac{6}{5}y=\frac{5}{4}$

$\bold{ y=-\frac{25}{24}}$

┈

$\rm x=-\frac{16}{5}\left(-\frac{25}{24}\right)+12$

$\rm x=\frac{10}{3}+12$

$\bold{ x=\frac{46}{3}}$

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In triangle IJK, J = 800cm, k = 240 cm and angle I=173 degrees. Find the area of triangle IJK, to the nearest square centimeter.

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