Respuesta :

One way to solve this is to use Pythagorean theorem: the square of one leg of triangle plus square of other leg of the triangle equals c the hypotenuse (longest side of triangle). You might see this as the formula a^2 + b^2 = c^2, where a and b are the legs and c is the hypotenuse.
In this case, the legs are 3√2 and the hypotenuse is h.
Using the formula:
(3√2)² + (3√2)² = h²
18 + 18 = h²
h = 6

The other way to do this is with trigonometric angles.
Remember cosine is adjacent over hypotenuse.
cos(45°) = (3√2) / h
h = (3√2) / cos(45°)
h = 6

The value hypotenuse of given triangle is [tex]\fbox{\bf 6 units}[/tex].

Further explanation:

Method 1:

Pythagoras theorem:

For a [tex]\triangle ABC[/tex], right angle at [tex]B[/tex] following relation holds:

[tex]{\left({AC}\right)^2}={\left( {AB}\right)^2}+{\left( {BC}\right)^2}[/tex]     ……(1)

Consider the given triangle as [tex]\triangle ABC[/tex] shown below in Figure 1.

Substitute [tex]3\sqrt 2[/tex]  for [tex]AB[/tex], [tex]3\sqrt 2[/tex] for [tex]BC[/tex] and [tex]h[/tex] for [tex]AC[/tex] in equation (1).

[tex]\begin{aligned}{h^2}&={\left( {3\sqrt2}\right)^2}+{\left({3\sqrt2}\right)^2}\\{h^2}&=18+18\\{h^2}&=36\\h&=6\\\end{aligned}[/tex]

Thus, the value hypotenuse of given triangle is [tex]\fbox{\bf 6 units}[/tex].

Method 2:

Consider the given triangle as [tex]\triangle ABC[/tex] as shown below in Figure 1.

In right [tex]\triangle ABC[/tex], [tex]BC[/tex] is base, [tex]AB[/tex] is perpendicular and [tex]AC[/tex] is hypotenuse.

[tex]sin[/tex] is defined as the ratio of perpendicular to hypotenuse, [tex]cos[/tex] is defined as the ratio of base to hypotenuse and [tex]tan[/tex] is defined as the ratio of perpendicular to base. There exists three more ratios which are reciprocal of [tex]sin,cos[/tex] and [tex]tan[/tex].

For [tex]\angle C[/tex], these ratios can be listed as shown below.

[tex]\begin{aligned}{\text{sinC=}}\frac{{AB}}{{AC}},\,{\text{\,\,\,\,\,\,\,\,cosecC=}}\frac{{AC}}{{AB}}\hfill\\{\text{cosC=}}\frac{{BC}}{{AC}},{\text{\,\,\,\,\,\,\,secC=}}\frac{{AC}}{{BC}}\hfill\\\tan C {\text{}}\frac{{AB}}{{BC}},\,\,\,\,\,\,\,\,\,\,\cot\,C{\text{=}}\frac{{BC}}{{AB}} \hfill\\\end{aligned}[/tex]

Substitute [tex]3\sqrt 2[/tex]  for [tex]AB[/tex], [tex]h[/tex] for [tex]AC[/tex] and 45 for [tex]\angle C[/tex].

[tex]\begin{aligned}{\text{sin45=}}&\frac{{3\sqrt2}}{h}\hfill\\\frac{1}{{\sqrt2}}&=\frac{{3\sqrt 2 }}{h}\hfill\\h&=\left({3\sqrt2}\right)\left({\sqrt2}\right)\hfill\\h&=6\hfill\\\end{aligned}[/tex]

Thus, the value hypotenuse of given triangle is [tex]\fbox{\bf 6 units}[/tex].

Learn more:  

1. Which rule describes the transformation? https://brainly.com/question/2992432  

2. Which undefined term is needed to define an angle? https://brainly.com/question/3717797  

3. Look at the figure, which trigonometric ratio should you use to find x? https://brainly.com/question/9880052  

Answer Details :

Grade: Senior School  

Subject: Mathematics  

Chapter: Triangle.

Keywords:

triangle ABC, sin, cos, tan, cot, sec, cosec, angle, Pythagoras theorem, hypotenuse, base, perpendicular, trigonometric ratio, similarity, ratio of sides, right triangle, similar triangle, ratio of sides, equal angles, square of hypotenuse, sum, square of legs, sum of square of legs, sum of angle of triangle, property of triangle, triangle ABC.  

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