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Free_Kalibri [48]
3 years ago
11

Can anyone please help me with this RSM question? I have literally posted this like four times!

Mathematics
2 answers:
Ostrovityanka [42]3 years ago
5 0

Answer:

Angles in a triangle add up to 180.

ABG - x + 90 + ANGLE AGB = 180

Angles on a straight add up to 180.

CDE - 8x + ANGLE D + 3x = 180

Alterate angles - ATTACHMENT

ANGLE D = x

8x + x + 3x = 180

12x = 180

x = 180/12

<h2>x = 15</h2>

zlopas [31]3 years ago
5 0

Answer:

x = 15

Step-by-step explanation:

Line FD is perpendicular to line EB. Line AC is perpendicular to line EB. Lines that are perpendicular to the same line are parallel. That makes lines FD and AC parallel. Angles EDF and C are corresponding angles of parallel lines cut by a transversal, so they are congruent. That makes m<C =  m<EDF = 3x.

The sum of the measures of the angles of a triangle is 180 degrees.

Look at triangle ACD.

m<A + m<C + m<ADC = 180

x + 3x + 8x = 180

12x = 180

x = 15

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\qquad \qquad\huge \underline{\boxed{\sf Answer}}

Here's the solution ~

Let's find the measure of hypotenuse first, by using Pythagoras theorem ;

\qquad \sf  \dashrightarrow \: h {}^{2}  =  {8}^{2}  +  {6}^{2}

\qquad \sf  \dashrightarrow \: h {}^{2}  =  {36}^{}  +  {64}^{}

\qquad \sf  \dashrightarrow \: h {}^{2}  = 100

\qquad \sf  \dashrightarrow \: h {}^{}  =  \sqrt{100}

\qquad \sf  \dashrightarrow \: h {}^{}  =  {10}

Now, let's find the asked values ~

\qquad \sf  \dashrightarrow \:  \sin(x) =  \dfrac{opposite \: side}{hypotenuse}

\qquad \sf  \dashrightarrow \:  \sin(x) =  \dfrac{6}{10}

\qquad \sf  \dashrightarrow \:  \sin(x) =  \dfrac{3}{5}   \: or \: 0.6 \: units

For Cos y :

\qquad \sf  \dashrightarrow \:  \cos(y) =  \dfrac{adjcant \: side}{hypotenuse}

\qquad \sf  \dashrightarrow \:  \cos(y)   = \dfrac{6}{10}

\qquad \sf  \dashrightarrow \:  \cos(y)   = \dfrac{3}{5}  \: or \: 0.6 \: units

As we can see that both sin x and Cos y have equal values, therefore The required relationships is equality.

I.e Sin x = Cos y

Hope it helps ~

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