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3 years ago

Help! Geometry question.

Mathematics

1 answer:

valentinak56 [21]3 years ago

6 0

Start with congruencies of 130°:

∠1 and 130° are vertical angles

∠2 and 130° are corresponding angles

∠5 and 130° are alternate exterior angles

m∠1 = m∠2 = m∠5 = 130°

∠7 and 130° are supplementary angles (which means their sum is 180°). So, ∠7 = 50°

Next, find congruencies of ∠7:

∠3 and ∠7 are vertical angles

∠4 and ∠7 are corresponding angles

∠6 and ∠7 are alternate exterior angles

m∠3 = m∠4 = m∠6 = m∠7 = 50°

Answer: B

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Show Work Please Thank You

Misha Larkins [42]

Answer:

$\displaystyle{x = \dfrac{\pi}{12}, \dfrac{\pi}{6}}$

Step-by-step explanation:

We are given the trigonometric equation of:

$\displaystyle{\sin 4x = \dfrac{\sqrt{3}}{2}}$

Let u = 4x then:

$\displaystyle{\sin u = \dfrac{\sqrt{3}}{2}}\\\\\displaystyle{\arcsin (\sin u) = \arcsin \left(\dfrac{\sqrt{3}}{2}\right)}\\\\\displaystyle{u= \arcsin \left(\dfrac{\sqrt{3}}{2}\right)}$

Find a measurement that makes sin(u) = √3/2 true within [0, π) which are u = 60° (π/3) and u = 120° (2π/3).

$\displaystyle{u = \dfrac{\pi}{3}, \dfrac{2\pi}{3}}$

Convert u-term back to 4x:

$\displaystyle{4x = \dfrac{\pi}{3}, \dfrac{2\pi}{3}}$

Divide both sides by 4:

$\displaystyle{x = \dfrac{\pi}{12}, \dfrac{\pi}{6}}$

The interval is given to be 0 ≤ 4x < π therefore the new interval is 0 ≤ x < π/4 and these solutions are valid since they are still in the interval.

Therefore:

$\displaystyle{x = \dfrac{\pi}{12}, \dfrac{\pi}{6}}$

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