# Suppose that x > 0 , cos(θ) = 45/x and sin(θ) = 24/x . Find x

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### 3 Answers

sin a = 45/ x

cos a = 24/x

We are given that sine and cosine of an angle.

Then we could use the trigonometric properties to solve:

We know that:

sin^2 a + cos^2 a = 1

Then we will substitute:

==> (45/x)^2 + (24/x)62 = 1

==> ( 2025/ x^2 + 576/ x^2 = 1

==> ( 2025 + 576) / x^2 = 1

==> (2601) / x^2 = 1

Now multiply by x^2:

==> x^2 = 2601

==> x = +-51

Then there are two solutions:

**x = { -51, 51}**

First, we'll impose the constraint of existence of the trigonometric functions sine and cosine:

-1=< sin(θ) =<1

sin(θ) = 24/x

-1=< 24/x =<1

We'll multiply both sides by x:

-x=< 24 =< x

-1=< cos(θ) =<1

cos(θ) = 45/x

-1=< cos(θ) =<1

-1=< 45/x =<1

We'll multiply both sides by x:

-x=< 45 =< x

From both inequalities, we'll get the interval for adissible value for x: [45 ; +infinite)

Now, we'll solve the equtaion, applying the fundamental formula of trigonomtery:

[sin(θ)]^2 + [cos(θ)]^2 =1

(576+2025)/x^2 = 1

We'll multiply both sides by x^2:

2601 = x^2

We'll apply square root both sides:

**x = 51**

**x = -51**

**Since just 51 is in the interval of admissible values, we'll accept just a single solution x = 51.**

For x>0, cost = 45/x and sin t = 24/x. To find x.

cost = 45/x and sint = 24/x.

Therefore cos^2t +sin^2t = (45/x)^2+(24/x)^2 .

But cos^2t+sin^2t = 1.

Therefore 45^2/x^2+24^2/x^2 = 1.

45^2+24^2 = x^2.

2025+576 = x^2.

2601 = x^2

x^2 = sqrt(2601) = 51, Or x = -51.

Therefore x = 51, as x>0 .

Th