First I am going to assume that these are both right triangles based off of look and because it is much easier. Without it you have to use law of sines or law of cosines...
So to find x you must first find y which can be done simply by using the pythagorean theorem. This theorem is defined as the sum of the squared legs is equal to the sum of the hypotenuse or x^2 + y^2 = z^2

If we substitute in the known values 16^2 + y^2 = 20^2 and solve for y we get that y = sqrt(20^2 - 16^2), this then simplifies to y = 12

Finding x is much more annoying, the easiest way I can immediately see is to find the upper angles by doing sin(16/20) and then 90 - sin(16/20) since the complementary angle is the one you want. I don't have a calculator or a trig table with me right now but I will tell you that x will be equal to 12 ÷ the inverse cosine of the angle (90degrees - sin(16/20)).

I am pretty sure the answer is D though because we know for sure y = 12 and x has to be greater than y because the hypotenuse must be larger than both legs. It could be E but you won't know unless you do the math for x. So it is either D or E but I would be surprised if a Professor made you do all of the work just to say it doesn't work...

8 0

3 years ago

Prove this trigonometric equation;<br><br> - tan^2x + sec^2x = 1,

alekssr [168]

Hey there :)

- tan²x + sec²x = 1 or 1 + tan²x = sec²x

sin²x + cos²x = 1
Divide the whole by cos²x

$\frac{sin^2x}{cos^2x} + \frac{cos^2x}{cos^2x} = \frac{1}{cos^2x}$

$\frac{sinx}{cosx} = tanx$ so $\frac{sin^2x}{cos^2x} = tan^2x$
and
$\frac{1}{cosx} = secx$ so $\frac{1}{cos^2x} = sec^2x$

Therefore,
tan²x + 1 = sec²x
Take tan²x to the other side {You will have the same answer}

1 = - tan²x = sec²x or sec²x - tanx = 1

3 0

3 years ago

Two in 1 more points

cluponka [151]

Answer:

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Step-by-step explanation:

4 0

3 years ago

Read 2 more answers