The distance between two pillars of length 16 metres and 9 metres is x metres. If two angles of elevation of..

their respective top from the bottom of the other are complementary to each other, then find the value of x.

tan(α) = 16/x
tan(β) = 9/x

α + β = π/2
tan(α + β) = tan(π/2)
1/tan(α + β) = 1/tan(π/2) = cot(π/2) = 0
(1 - tan(α) tan(β)) / (tan(α) + tan(β)) = 0
1 - tan(α) tan(β) = 0
1 - (16/x) (9/x) = 0
1 = 144/x²
x² = 144
x = 12

Mαthmφm

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I'll also include my answer to a previous question you asked. For some reason, it got deleted before I could post my answer.

Your question:

If cos^2(x) + cos^4 (x) = 1 when x is a positive acute angle, then the value of tan^2(x) + tan^4 (x) is?
a. 3/2
b. 1
c. 1/2
d. 0

Please explain how ?

My answer:

cos²(x) + cos⁴(x) = 1
cos⁴(x) + cos²(x) - 1 = 0

Use quadratic formula to solve for cos²(x)

cos²(x) = (-1 ± √(1+4)) / 2 ------> disregard negative value

cos²(x) = (√5 - 1) / 2

sin²(x) = 1 - cos²(x) = 1 - (√5 - 1) / 2 = (3 - √5) / 2

tan²(x) = sin²(x)/cos²(x)
. . . . . = (3 - √5) / (√5 - 1)
. . . . . = (3 - √5)(√5 + 1) / ((√5 - 1)(√5 + 1))
. . . . . = (3√5 + 3 - 5 - √5) / (5 - 1)
. . . . . = (2√5 -2) / 4
. . . . . = (√5 - 1) / 2

tan⁴(x) = ((√5 - 1) / 2)²
. . . . . = (√5 - 1)(√5 - 1) / 4
. . . . . = (5 - √5 - √5 + 1) / 4
. . . . . = (6 - 2√5) / 4
. . . . . = (3 - √5) / 2

tan²(x) + tan⁴(x)
= (√5 - 1) / 2 + (3 - √5) / 2
= (√5 - 1 + 3 - √5) / 2
= 2 / 2
= 1

Mαthmφm

x = 12m
Here's a picture of it with the solution:
http://img197.imageshack.us/img197/337/m…