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2 years ago

Which triangle is congruent to triangle PQR

Mathematics

1 answer:

alexgriva [62]2 years ago

4 0

Answer:

Triangular PQR is congruent to triangular ABC

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Solve the equation for all real values of x.<br> cosxtanx - 2 cos x=-1

IceJOKER [234]

Solution to equation $cosxtanx - 2 cos^2 x=-1$ for all real values of x is $x=2k\pi + \frac{\pi}{6} , x=2k\pi + \frac{5\pi}{6}$ .

<u>Step-by-step explanation:</u>

Here we have , $cosxtanx - 2 cos^2 x=-1$ . Let's solve :

⇒ $cosxtanx - 2 cos^2 x=-1$

⇒ $cosx(\frac{sinx}{cosx}) - 2 cos^2 x=-1$

⇒ $sinx = 2 cos^2 x-1$

⇒ $sinx = 2 (1-sin^2x)-1$

⇒ $sinx = 1-2sin^2x$

⇒ $2sin^2x+sinx-1=0$

By quadratic formula :

⇒ $sinx = \frac{-b \pm \sqrt{b^2-4ac} }{2a}$

⇒ $sinx = \frac{-1 \pm \sqrt{1^2-4(2)(-1)} }{2(2)}$

⇒ $sinx = \frac{-1 \pm3}{4}$

⇒ $sinx = \frac{1}{2} , sinx =-1$

⇒ $sinx = sin\frac{\pi}{6} , sinx = sin\frac{3\pi}{2}$

⇒ $x=\frac{\pi}{6} , x=\frac{3\pi}{2}$

But at $x=\frac{3\pi}{2}$ we have equation undefined as $cos\frac{3\pi}{2}=0$ . Hence only solution is :

⇒ $x=\frac{\pi}{6}$

Since , $sin(\pi -x)=sinx$

⇒ $x=\pi -\frac{\pi}{6} = \frac{5\pi}{6}$

Now , General Solution is given by :

⇒ $x=2k\pi + \frac{\pi}{6} , x=2k\pi + \frac{5\pi}{6}$

Therefore , Solution to equation $cosxtanx - 2 cos^2 x=-1$ for all real values of x is $x=2k\pi + \frac{\pi}{6} , x=2k\pi + \frac{5\pi}{6}$ .

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How to find the θ in this equation: tan(θ+16°)=1/tan(θ+60°) by using the trigonometric identities?thanks

DedPeter [7]

Answer:

$\huge \theta =7°$

Step-by-step explanation:

$tan( \theta + 16 \degree) = \frac{1}{tan( \theta + 60 \degree) } \\ tan( \theta + 16 \degree) =cot( \theta + 60 \degree) \\ tan( \theta + 16 \degree) = tan \{90 \degree - (\theta + 60\degree) \} \\ \theta + 16 \degree = 90 \degree - \theta - 60\degree \\ \theta +\theta = 30 \degree - 16\degree \\2 \theta = 14 \degree \\ \theta = \frac{14 \degree}{2} \\ \huge \red{ \boxed{\theta =7 \degree}}$

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