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2 years ago

What is the answer to this question?

Mathematics

1 answer:

AleksAgata [21]2 years ago

6 0

Answer:

acute and iscoles

Step-by-step explanation:

acute is >90

its not right or obtuse so its acute. so its either scalene on iscoloses

scalene is 3 unequal sides but as you can see it has 2 sides that are equal so it cant be scalene. Therefore its iscolese

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3 years ago

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3 years ago

If <img src="https://tex.z-dn.net/?f=tan%20%28x%29%20%3D%20%5Cfrac%7B5%7D%7B12%7D" id="TexFormula1" title="tan (x) = \frac{5}{12

Alekssandra [29.7K]

Explanation:

First, we need to find the values of the sine and cosine of x knowing the value of tan x and x being in the 3rd quadrant. Since tan x = 5/12, using Pythagorean theorem, we know that

$\sin x = -\frac{5}{13}\;\;\text{and}\;\;\cos x = -\frac{12}{13}$

Note that both sine and cosine are negative because x is in the 3rd quadrant.

Recall the addition identities listed below:

$\sin(\alpha + \beta) = \sin\alpha\sin\beta + \cos\alpha\cos\beta$

$\Rightarrow \sin(180+x) = \sin180\sin x + \cos180\cos x$

$\;\;\;\;\;\;= -\sin x = \dfrac{5}{13}$

$\cos(\alpha - \beta) = \cos\alpha \cos\beta + \sin\alpha \sin\beta$

$\Rightarrow \cos(180 - x) = \cos180\cos x + \sin180\sin x$

$\;\;\;\;\;\;=-\cos x = \dfrac{12}{13}$

$\tan(\alpha - \beta) = \dfrac{\tan\alpha - \tan\beta}{1 + \tan\alpha\tan\beta}$

$\Rightarrow \tan(360 - x) = \dfrac{\tan 360 - \tan x}{1 + \tan 360 \tan x}$

$\;\;\;\;\;\;= -\tan x = -\dfrac{5}{12}$

Therefore, the expression reduces to

$\sin(180+x) + \tan(360-x) + \frac{1}{\cos(180-x)}$

$\;\;\;\;\;= \left(\dfrac{5}{13}\right) + \left(\dfrac{5}{12}\right) + \dfrac{1}{\left(\frac{12}{13}\right)}$

$\;\;\;\;\;= \dfrac{49}{26}$

5 0

2 years ago