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4 votes
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

2 Answers

3 votes

Answer:


x = 59.16 \degree

Explanation:

We are given a right angled triangle, with two sides as 4.9 and 8 . And we are interested in finding out the measure of the angle "x" . With respect to angle x , the side measuring 8 is perpendicular and that measuring 4.9 is base.

In a right angled triangle, tangent is defined as the ratio of perpendicular and base .

So , according to the question,

  • p = 8
  • b = 4.9

Now we have;


\implies \tan\theta =(p)/(b) \\

Here the angle is " x " , so that ,


\implies \tan x =(8)/(4.9)\\

Simplify,


\implies \tan x = 1.63 \\

Take arctan on both the sides,


\implies \arctan (\tan x ) = \arctan(1.63) \\

Simplify,


\implies x = \arctan(1.63)\\


\implies \underline{\underline{ x = 59.16\degree \approx 59.2\degree}}\\

Hence the value of x is 59.2° .

answered
User River
by
8.0k points
7 votes

Answer:

x ≈ 58.5°

Explanation:

using the tangent ratio in the right triangle

tan x =
(opposite)/(adjacent) =
(DE)/(EF) =
(8)/(4.9) , then

x =
tan^(-1) (
(8)/(4.9) ) ≈ 58.5° ( to the nearest tenth )

answered
User Ximbal
by
8.3k points

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