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2 years ago

Three sides of a triangle measure 10m, 12m, and 18m. Find the largest angle of the triangle to the nearest degree.

Mathematics

1 answer:

sweet-ann [11.9K]2 years ago

8 0

Answer:

109°

Step-by-step explanation:

The Law of Cosines is good for finding angles when you know three sides. It tells you ...

a^2 = b^2 + c^2 -2bc·cos(A)

where a, b, c are the side lengths and A, B, C are the opposite angles. Here, we want a=18, since the longest side is opposite the largest angle.

Solving for cos(A), we get ...

(b^2 +c^2 -a^2)/(2bc) = cos(A)

A = arccos((b^2 +c^2 -a^2)/(2bc)) = arccos((10^2 +12^2 -18^2)/(2·10·12))

= arccos(-80/240) = arccos(-1/3) ≈ 109.471°

The largest angle of the triangle is about 109°.

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<h2>Explanation:</h2>

The picture of the garden is shown below. In this problem, we have:

One square
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Recall that the area of a square is given by:

$A_{s}=L^2 \\ \\ A_{s}:Area \ of \ a \ square \\ L:Side \ of \ the \ square$

On the other hand, the area of a rectangle is given by:

$A_{r}=L_{1}\times L_{2} \\ \\ \\ A_{r}:Area \ of \ a \ rectangle \\ \\ L_{1}:Side \ 1 \ of \ the \ rectangle \\ \\ L_{2}:Side \ 2 \ of \ the \ rectangle$

Therefore:

FOR THE SQUARE:

$L=4m \\ \\ \\ Substituting \ values: \\ \\ \\ A_{s}=4^2 \\ \\ \boxed{A_{s}=16m^2}$

FOR THE RECTANGLE 1:

$L_{1}=6m \\ L_{2}=4m \\ \\ \\ Substituting \ values: \\ \\ \\ A_{r_{1}}=6(4) \\ \\ \boxed{A_{r_{1}}=24m^2}$

FOR THE RECTANGLE 2:

$L_{1}=10m \\ L_{2}=8m \\ \\ \\ Substituting \ values: \\ \\ \\ A_{r_{2}}=10(8) \\ \\ \boxed{A_{r_{2}}=80m^2}$

Finally, the area of a garden is the sum of the three areas:

$A_{total}=A_{s}+A_{r_{1}}+A_{r_{2}} \\ \\ A_{total}=16m^2+24m^2+80m^2 \\ \\ A_{total}=16+24+80 \\ \\ \boxed{A_{total}=120m^2}$

<h2>Learn more:</h2>

Area of a pizza: brainly.com/question/12878495

#LearnWithBrainly

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