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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 class=

Sagot :

Answer:

x = 33.4

Step-by-step explanation:

The diagram shows right triangle ABC, with the angle at vertex A marked as 'x', and side lengths BC = 3.3 and AC = 6.

As we have been given the lengths of the side opposite angle x and the hypotenuse, we can use the sine trigonometric ratio to find the value of x.

[tex]\boxed{\begin{array}{l}\underline{\textsf{Sine trigonometric ratio}}\\\\\sf \sin(\theta)=\dfrac{O}{H}\\\\\textsf{where:}\\\phantom{ww}\bullet\;\textsf{$\theta$ is the angle.}\\\phantom{ww}\bullet\;\textsf{O is the side opposite the angle.}\\\phantom{ww}\bullet\;\textsf{H is the hypotenuse (the side opposite the right angle).}\end{array}}[/tex]

In this case:

  • θ = x°
  • O = BC = 3.3
  • H = AC = 6

Substitute the values into the sine ratio and solve for x:

[tex]\sin x^{\circ}=\dfrac{3.3}{6} \\\\\\ \sin x^{\circ}=0.55 \\\\\\ x^{\circ}=\sin^{-1}(0.55) \\\\\\ x^{\circ}=33.36701296923...^{\circ}\\\\\\ x=33.4\; \sf (nearest\;tenth)[/tex]

Therefore, the value of x rounded to the nearest tenth is:

[tex]\LARGE\boxed{\boxed{x=33.4}}[/tex]

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