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3 years ago

Angle DEF is a right triangle. If FE = 12 and DE = 5, find DF.

Mathematics

2 answers:

tensa zangetsu [6.8K]3 years ago

6 0

Answer:

Option D is correct.

side DF = 13 units

Step-by-step explanation:

Using Pythagoras theorem:

$\text{Hypotenuse side}^2=\text{Opposite side}^2+\text{Adjacent side}^2$ ....[1]

In a right angle triangle DEF:

FE = 12 units and DE = 5 units.

Using Pythagoras theorem on DEF we have;

Hypotenuse side = DF

Opposite side = DE and Adjacent side = FE

Substitute these in [1] we have;

$DF^2 = 5^2+12^2 = 25 +144 = 169$

⇒ $DF = \sqrt{169} = 13$ units.

Therefore, the side DF = 13 units

Usimov [2.4K]3 years ago

3 0

c^2=a^2+b^2

c^2 = 12^2 + 5^2

C^2 = 144 +25

C^2 = 169

c=13

DF = 13

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Maths trig functions <br> please help

Nikitich [7]

1. According to the plots, the curves intersect when $x=90^\circ$ and $x=360^\circ$ .

We can confirm this algebraically.

$\sin(x) + 2 = -\cos(x) + 3$

$\sin(x) + \cos(x) = 1$

$\sqrt2 \sin\left(x + 45^\circ\right) = 1$

$\sin\left(x + 45^\circ\right) = \dfrac1{\sqrt2}$

$x + 45^\circ = \sin^{-1}\left(\dfrac1{\sqrt2}\right) + 360^\circ n \text{ or } x + 45^\circ = 180^\circ - \sin^{-1}\left(\dfrac1{\sqrt2}\right) + 360^\circ n$

(where $n$ is an integer)

$x + 45^\circ = 45^\circ + 360^\circ n \text{ or } x + 45^\circ = 135^\circ + 360^\circ n$

$x = 360^\circ n \text{ or } x = 90^\circ + 360^\circ n$

We get the two solutions we found in the interval [0°, 360°] with $n=1$ in the first case, and $n=0$ in the second case.

2. We have $\sin(x)=0$ when $x \in \{0^\circ, \pm 180^\circ, \pm360^\circ, \ldots\}$ . For the given plot domain [0°, 360°], this happens when $180^\circ < x < 360^\circ$ .