2 years ago

CAN SOMEONE HELP ME WITH THIS GEOMETRY QUESTION PLEASE?!?

1 answer:

4 0

The triangles are similar
so
(x+8)/42 = 24/(2x-4)
(x+8)(2x-4) = 24 * 42
2x^2 + 16x - 4x - 32 = 1008
2x^2 + 12x - 32 = 1008
2x^2 + 12x - 1040 = 0
2(x^2 + 6x - 520) = 0
2(x + 26)(x - 20) = 0
x + 26 = 0; x = -26
x - 20 = 0; x = 20

Since the side of triangle can't be negative so x = 20

Answer
x = 20

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2 years ago

Given that cos(theta) = 8/17 and that theta lies in Quadrant IV, what is the exact value of sin 2(theta)?

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Answer: $-\frac{240}{289}$

=========================================================

Explanation:

Use the pythagorean trig identity $\sin^2(\theta)+\cos^2(\theta) = 1$ and plug in the fact that $\cos(\theta) = \frac{8}{17}\\\\$

Isolating sine leads to $\sin(\theta) = -\frac{15}{17}\\\\$ . I'm skipping the steps here, but let me know if you need to see them.

The result is negative because we're in quadrant 4, when y < 0 so it's when sine is negative.

Therefore,

$\sin(2\theta) = 2\sin(\theta)\cos(\theta)\\\\\sin(2\theta) = 2*\left(-\frac{15}{17}\right)*\left(\frac{8}{17}\right)\\\\\sin(2\theta) = -\frac{240}{289}\\\\$