Trigonometry problem

Question

Suppose a= 10 and c=square root of 120.

find the exact value for sin(A), cos(A), tan(A), cot(A), sec(A), csc(A)

Answer 1

Right triangle ABC

A on the right, C bottom left, B top left

GIVEN is a = BC = 10 and

c = AB = sqrt(120) = sqrt(2*2*2*3*5)=2*sqrt(30)

Pythagorean says:

c^2 - a^2 = b^2

120 - 100 = b^2

20 = b^2

2*sqrt(5) = b = AC

Then sin A = 10 / (2*sqrt(30))

= 5/sqrt(30)

= 5*sqrt(30)/30 <--- rationalizes

= sqrt(30)/6

cos A = 2*sqrt(5) / 2 * sqrt(30)

= sqrT(5)/sqrt(30) <-- 2 cancels

= sqrt(5/30) =

sqrt(1/6) =

sqrt(6)/6 <--- rationalizes

tan A = 10 / [2*sqrt(5)]

= 5 / sqrt(5)

= sqrt(5) <-- by definition: opposite over adjacent

Note that tan = sin/cos, so substituting the above...

sin/cos = sqrt(30)/6 divided by sqrt(6)/6

= sqrt(30)/6 * 6 / sqrt(6) <-- KFC: keep ,change ,flip

= sqrt(30)/sqrt(6)

= sqrt(30/6)

= sqrt(5), which agrees with the previous result by definition

cosecant is then 6/sqrt(30) = 6*sqrt(30)/30 = sqrt(30)/5

secant is 6/sqrt(6) = sqrt(6)

and cotangent is sqrt(5)/5

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Now, for angle B, sin B = cos A = sqrt(6)/6

and [vice versa], so cos B = sin A = sqrt(30)/6

Then tan B = sin B/cos B = sqrt(6)/6 divided by sqrt(30)/6

= sqrt(6)/6 * 6/sqrt(30)

= sqrt(6)/sqrt(30)

= sqrt(6/30)

= sqrt(1/5)

= sqrt(5)/5

the csc B = sqrt(6)

sec B = 6/sqrt(30) = 6 * sqrt(30)/30 = sqrt(30)/5

cot B = sqrt(5)