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Suppose you have a right triangle called [tex]\triangle ABC[/tex] where [tex]\angle B[/tex] is the right angle. If [tex]\overline{AB} = 3\sqrt{5}[/tex] and [tex]\overline{AC} = 6\sqrt{7}[/tex], which expression can be used to find [tex]m\angle C[/tex]?

A. [tex]\sin^{-1}\left(\frac{\sqrt{35}}{14}\right)[/tex]
B. [tex]\sin^{-1}\left(\frac{2\sqrt{35}}{14}\right)[/tex]
C. [tex]\cos^{-1}\left(\frac{2\sqrt{35}}{14}\right)[/tex]
D. [tex]\cos^{-1}\left(\frac{\sqrt{35}}{14}\right)[/tex]

Sagot :

GinnyAnsweridn
To find the measure of angle [tex]\( \angle C \)[/tex] in the right triangle [tex]\( \triangle ABC \)[/tex] where [tex]\(\angle B\)[/tex] is the right angle, and we know the lengths of the sides [tex]\( \overline{AB} = 3\sqrt{5} \)[/tex] and [tex]\( \overline{AC} = 6\sqrt{7} \)[/tex], we can use trigonometric ratios and properties of right triangles. Here is the step-by-step solution:

1. Given Information:
- [tex]\( \overline{AB} = 3\sqrt{5} \)[/tex]
- [tex]\( \overline{AC} = 6\sqrt{7} \)[/tex]
- [tex]\( \angle B = 90^\circ \)[/tex]

2. Finding the Length of [tex]\( \overline{BC} \)[/tex]:
Since [tex]\(\triangle ABC\)[/tex] is a right triangle with [tex]\(\angle B = 90^\circ\)[/tex], we can use the Pythagorean theorem to find [tex]\( \overline{BC} \)[/tex] (hypotenuse [tex]\( \overline{AC} \)[/tex] and legs [tex]\( \overline{AB} \)[/tex] and [tex]\( \overline{BC} \)[/tex]):
[tex]\[ \overline{BC}^2 = \overline{AC}^2 - \overline{AB}^2 \][/tex]
[tex]\[ \overline{BC}^2 = (6\sqrt{7})^2 - (3\sqrt{5})^2 \][/tex]
[tex]\[ \overline{BC}^2 = 252 - 45 \][/tex]
[tex]\[ \overline{BC}^2 = 207 \][/tex]
[tex]\[ \overline{BC} = \sqrt{207} \][/tex]

3. Finding [tex]\( \sin(\angle C) \)[/tex]:
By definition, [tex]\( \sin(\angle C) = \frac{\text{opposite}}{\text{hypotenuse}} \)[/tex]. Here, [tex]\(\overline{AB}\)[/tex] is the side opposite to [tex]\(\angle C\)[/tex], and [tex]\(\overline{AC}\)[/tex] is the hypotenuse.
[tex]\[ \sin(\angle C) = \frac{3\sqrt{5}}{6\sqrt{7}} \][/tex]
Simplify the fraction:
[tex]\[ \sin(\angle C) = \frac{3\sqrt{5}}{6\sqrt{7}} = \frac{\sqrt{5}}{2\sqrt{7}} = \frac{\sqrt{5}}{2} \cdot \frac{1}{\sqrt{7}} = \frac{\sqrt{5}}{2\sqrt{7}} = \frac{\sqrt{35}}{14} \][/tex]

4. Finding [tex]\( \angle C \)[/tex]:
The angle [tex]\( \angle C \)[/tex] can be found by taking the inverse sine (arcsin) of [tex]\(\sin(\angle C)\)[/tex]:
[tex]\[ \angle C = \sin^{-1}\left( \frac{\sqrt{35}}{14} \right) \][/tex]

So, the correct expression to find [tex]\( m \angle C \)[/tex] is:

[tex]\[ \boxed{\sin^{-1}\left( \frac{\sqrt{35}}{14} \right)} \][/tex]

After calculation and checking the values and units, this solution directly corresponds to the first option given in the question.
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