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In ∆JKL, j=21 cm, < J=39° and

User BlackWasp
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1 Answer

6 votes

The sketch of the diagram is given above

Given: j = 21 cm

and


\begin{gathered} Since we have a triangle, the total angle in a triangle is 180 degrees[tex]\begin{gathered} <p>Having found all the angles, we will then use trigonometric identities to get the lengths of the other sides</p><p></p><p>We can use the sine rule</p><p></p>[tex]\begin{gathered} (\sin39^0)/(21)\text{ = }(\sin 118^0)/(l) \\ \\ To\text{ get l, we will cross multiply} \\ l\text{ = }\frac{\sin 118^0\text{ x 21}}{\sin 39^0} \\ \end{gathered}


\begin{gathered} l\text{ = }(18.5419)/(0.6293) \\ \\ l\text{ = 29.46 cm} \end{gathered}

To get the area, we can use the relation


\text{Area = }(1)/(2)\text{ x l x j x sin 23 }

we have established that l = 29.46cm, j = 21cm


\text{Area = }(1)/(2)\text{ x 21 x 29.46 x 0.3907}

Area = 120.86 square centimeter

=> 120.9 square centimeters to the nearest tenth

In ∆JKL, j=21 cm, < J=39° and-example-1
User Adnan Zameer
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