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Dominik [7]
3 years ago
8

If cos x= sin(20 + x)° and 0° < x < 90°, what is the value of x

Mathematics
1 answer:
ZanzabumX [31]3 years ago
3 0

If cos x= sin(20 + x)° and 0° < x < 90° then value of x is 35 degrees

<em><u>Solution:</u></em>

Given that:

cos x= sin (20 + x)

We know that,

sin (a+b)=sin a \times cos b+cos a \times sin b

cos x= sin (20 + x) = sin 20 cos x + cos 20 sin x

cos x-sin20 \times cos x=cos 20 \times sin x

Taking cos x as common,

cos x(1-sin 20)=cos 20 \times sin x

\frac{(1-sin 20)}{(cos 20)}=\frac{sin x}{cos x }\\\\tan x = \frac{(1-sin 20)}{(cos 20)}

By trignometric functions,

sin 20 = 0.34202

cos 20 = 0.939692

So,

tan x = \frac{1 - 0.34202}{0.939692}\\\\tan x = 0.7002

Therefore,

x = arc tan (0.7002)

x = 35 degrees

Therefore value of x is 35 degrees

<h3><u>Method 2:</u></h3>

cos x = sin (20 + x)

sin and cos are co - functions, which means that:

cos x = cos [90 - (20 + x)]

cos x = cos (90 - 20 - x)

cos x = cos (70 - x)

Therefore, x = 70 - x

x + x = 70

2x = 70

x = 35

Therefore value of x is 35 degrees

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Step-by-step explanation:

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                  4*cos(Q) + 2.6*cos(Q) = u_s*4*sin(Q)

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