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Solve for x. Round to the nearest tenth of a degree, if necessary.

Solve for x. Round to the nearest tenth of a degree, if necessary.-example-1
User Elon Zito
by
8.0k points

2 Answers

3 votes

Answer:

x = 48.6°

Explanation:

We are heree given a right angled triangle in which we are interested in finding the value of unknown angle " x " . with respect to angle x , HG is perpendicular and FH is hypotenuse. Also , in right angled triangle, sine is defined as the ratio of perpendicular and hypotenuse.

So that ,


\implies \sin\theta = (HG)/(HF) \\


\implies \sin x =(1.2)/(1.6)\\


\implies \sin x = (3)/(4)\\


\implies x = \arcsin\bigg((3)/(4)\bigg)\\


\implies x = 48.59^o\\

Rounding to nearest tenth,


\implies x = 48.6^o \\

Hence the value of x is 48.6° .

User Bivek
by
8.6k points
2 votes

Answer:

48.6

Explanation:

SohCahToa

Using Sin^-1

Sin^-1 (1.2/1.6) = 48.5903

User Dabrut
by
8.4k points

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