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In ΔIJK, i = 340 cm, ∠I=120° and ∠J=59°. Find the area of ΔIJK, to the nearest 10th of a square centimeter.

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shisiraryal

Answer:

8900 is the correct answer

MrRoyal

The area of triangle IJK is 999.8 square centimeters

The given parameters are:

[tex]\angle I = 120^o[/tex]

[tex]\angle J = 59^o[/tex]

[tex]i = 340cm[/tex]

Start by calculating the length of side j using the following sine ratio

[tex]\frac{i}{\sin(I)} = \frac{j}{\sin(J)}[/tex]

Make j the subject of the formula

[tex]j = \sin(J) \times \frac{i}{\sin(I)}[/tex]

Substitute known values

[tex]j = \sin(59) \times \frac{340}{\sin(120)}[/tex]

Evaluate the expression

[tex]j = 337[/tex]

Next, calculate angle K using the following sum of angles formula

[tex]\angle I + \angle J + \angle K =180[/tex]

Substitute known values

[tex]120 + 59 + \angle K =180[/tex]

[tex]179 + \angle K =180[/tex]

Subtract 179 from both sides

[tex]\angle K =1[/tex]

The area of the triangle is:

[tex]Area = \frac{1}{2} ij \times \sin(K)[/tex]

So, we have:

[tex]Area = \frac{1}{2} \times 340 \times 337 \times \sin(1)[/tex]

Evaluate

[tex]Area = 999.8[/tex]

Hence, the area of triangle IJK is 999.8 square centimeters

Read more about areas at:

https://brainly.com/question/2391510

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