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3 years ago

In triangle JKL, solve for x

Mathematics

2 answers:

Archy [21]3 years ago

8 0

For this question you need to apply the trigonometric ratios with cos 27 degrees
cos 27 = $\frac{34}{x}$
0.891x = 34cm
∴x = 38.159cm

please mark as brainliest

Alchen [17]3 years ago

3 0

38.16 = when rounded

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6 0

1 year ago

PLEASE MATH HELP ( EASY )

Natalija [7]

Answer:

Step-by-step explanation:

28° into radians

28° = ( 28 * $\frac{\pi }{180}$ )

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= ( $\frac{7\pi }{45}$ )

Hope it helped :)

6 0

3 years ago

Read 2 more answers

A rectangle has width that is 6 meters less than the length. The area of the rectangle is 280 square meters. Find the dimensions

salantis [7]

Dimensions are length 20 meter and width 14 meter

Solution:

Let "a" be the length of rectangle

Let "b" be the width of rectangle

Given that,

A rectangle has width that is 6 meters less than the length

Width = length - 6

b = a - 6

The area of the rectangle is 280 square meters

The area of the rectangle is given by formula:

$Area = length \times width$

Substituting the values we get,

$Area = a \times (a-6)\\\\280 = a^2-6a\\\\a^2-6a -280=0$

Solve the above equation by quadratic formula

$\mathrm{For\:a\:quadratic\:equation\:of\:the\:form\:}ax^2+bx+c=0\mathrm{\:the\:solutions\:are\:}\\\\\quad x_{1,\:2}=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

$\mathrm{For\:}\quad a=1,\:b=-6,\:c=-280:\quad a_{1,\:2}=\frac{-\left(-6\right)\pm \sqrt{\left(-6\right)^2-4\cdot \:1\left(-280\right)}}{2\cdot \:1}$

$a =\frac{6 \pm \sqrt{36+1120}}{2}\\\\a = \frac{6 \pm \sqrt{1156}}{2}\\\\a = \frac{6 \pm 34}{2}\\\\Thus\ we\ have\ two\ solutions\\\\a = \frac{6+34}{2} \text{ or } a = \frac{6-34}{2}\\\\a = 20 \text{ or } a = -14$