If $\alpha$ is an angle in a triangle and if $\tan\alpha = 7$, then which of the statements is true?

Question

If $\alpha$ is an angle in a triangle and if $\tan\alpha = 7$, then which of the following statements is true?

a) $0<\alpha<\frac{\pi}{6}$

b) $\frac{\pi}{3}<\alpha<\frac{\pi}{2}$

c) There exists no such angle.

d) None of the above.

Since $$\tan\alpha = \frac{\sin\alpha}{\cos\alpha}=7,$$ both $\sin\alpha$ and $\cos\alpha$ should be either negative or positive. This only happens in 1° and 3° quadrant. So both option a) and b) should satisfy this equation since both lie in the first quadrant. But there is only one correct answer.

Answer 1

You need to consider when $\tan(\alpha)>1,$ so this is the same as asking when is $\dfrac{\sin\alpha}{\cos\alpha}>1.$ Once you find that out, you ask, when is $\tan(x)$ undefined. Your answer lies in between those values.

Answer 2

(1) $\tan{0}=0$

(2) $\tan{\pi/6}=1/\sqrt{3}$

(3) $\tan$ monotonically increases on $[0,\pi/6]$.

Answer 3

Hint:

$$\tan\left(\frac\pi3\right)=\sqrt3<7<+\infty=\lim_{\alpha\to\pi/2^-}\tan\left(\alpha\right)$$

Answer 4

hint

We have $$\tan (\frac {\pi}{3})=\sqrt {3} $$ and $$\tan(\frac {\pi}{3+3})=\frac{\sqrt {3}}{3} <\sqrt {3}<7$$

and $x\mapsto \tan (x) $ increasing at $[0,\frac {\pi}{2}) $.

$$\frac {\pi}{3}<\alpha <\frac {\pi}{2} $$

Answer 5

You continue as follows:

$$\tan{\alpha}=7 \Rightarrow \frac{\sin{\alpha}}{\cos{\alpha}}=7 \Rightarrow \sin{\alpha}=7\sqrt{1-\sin^2{\alpha}}$$

Noting $0<\alpha<\frac{\pi}{2},$ we square both sides: $$ \sin^2{\alpha}=49(1-\sin^2{\alpha}) \Rightarrow $$ $$\sin^2{\alpha}=\frac{49}{50} \Rightarrow\sin{\alpha}=\frac{7}{5\sqrt{2}}\approx 0.99 \Rightarrow \frac{\pi}{3}<\alpha<\frac{\pi}{2}.$$

Note: $\sin{\alpha}$ increases in $0<\alpha<\frac{\pi}{2}$ and $\sin{90^\circ}=1.$