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3 years ago

FAST!! Evaluate tan60/cos45 √6 √3/2 √2/3 1√6

Mathematics

2 answers:

evablogger [386]3 years ago

4 0

Answer:

$\frac{\tan 60\degree}{\cos45 \degree}= \sqrt{6}$

Step-by-step explanation:

We want to evaluate

$\frac{\tan 60\degree}{\cos45 \degree}$

We use special angles or the unit circle to obtain;

$\frac{\tan 60\degree}{\cos45 \degree}=\frac{\sqrt{3}}{\frac{\sqrt{2}}{2}}$

This implies that;

$\frac{\tan 60\degree}{\cos45 \degree}=\sqrt{3}\div \frac{\sqrt{2}}{2}$

$\frac{\tan 60\degree}{\cos45 \degree}=\sqrt{3}\times \sqrt{2}$

$\frac{\tan 60\degree}{\cos45 \degree}= \sqrt{6}$

Leokris [45]3 years ago

3 0

Answer:

$\sqrt{6}$ .

Step-by-step explanation:

$\frac{tan(60)}{cos(45)}$

$= \frac{\frac{sin(60)}{cos(60)}}{cos(45)}$

$= \frac{sin(60)}{cos(60)*cos(45)}$

$= \frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}*\frac{\sqrt{2}}{2}}$

$= \frac{\frac{\sqrt{3}}{2}}{\frac{\sqrt{2}}{4}}$

$= \frac{4\sqrt{3}}{2\sqrt{2}}$

$= \frac{2\sqrt{3}}{\sqrt{2}}$

$= \frac{2\sqrt{3}\sqrt{2}}{2}$

$=\sqrt{3}\sqrt{2}$

$=\sqrt{6}$ .

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Two lighthouses are located 75 miles from one another on a north-south line. If a boat is spotted S 40o E from the northern ligh

yuradex [85]

Answer:

The northern lighthouse is approximately $24.4\; \rm mi$ closer to the boat than the southern lighthouse.

Step-by-step explanation:

Refer to the diagram attached. Denote the northern lighthouse as $\rm N$ , the southern lighthouse as $\rm S$ , and the boat as $\rm B$ . These three points would form a triangle.

It is given that two of the angles of this triangle measure $40^{\circ}$ (northern lighthouse, $\angle {\rm N}$ ) and $21^{\circ}$ (southern lighthouse $\angle {\rm S}$ ), respectively. The three angles of any triangle add up to $180^{\circ}$ . Therefore, the third angle of this triangle would measure $180^{\circ} - (40^{\circ} + 21^{\circ}) = 119^{\circ}$ (boat $\angle {\rm B}$ .)

It is also given that the length between the two lighthouses (length of $\rm NS$ ) is $75\; \rm mi$ .

By the law of sine, the length of a side in a given triangle would be proportional to the angle opposite to that side. For example, in the triangle in this question, $\angle {\rm B}$ is opposite to side $\rm NS$ , whereas $\angle {\rm S}$ is opposite to side ${\rm NB}$ . Therefore:

$\begin{aligned} \frac{\text{length of NS}}{\sin(\angle {\rm B})} = \frac{\text{length of NB}}{\sin(\angle {\rm S})} \end{aligned}$ .

Substitute in the known measurements:

$\begin{aligned} \frac{75\; \rm mi}{\sin(119^{\circ})} = \frac{\text{length of NB}}{\sin(21^{\circ})} \end{aligned}$ .

Rearrange and solve for the length of $\rm NB$ :

$\begin{aligned} & \text{length of NB} \\ =\; & (75\; \rm mi) \times \frac{\sin(21^{\circ})}{\sin(119^{\circ})} \\ \approx\; & 30.73\; \rm mi\end{aligned}$ .

(Round to at least one more decimal places than the values in the choices.)

Likewise, with $\angle {\rm N}$ is opposite to side ${\rm SB}$ , the following would also hold:

$\begin{aligned} \frac{\text{length of NS}}{\sin(\angle {\rm B})} = \frac{\text{length of SB}}{\sin(\angle {\rm N})} \end{aligned}$ .

$\begin{aligned} \frac{75\; \rm mi}{\sin(119^{\circ})} = \frac{\text{length of SB}}{\sin(40^{\circ})} \end{aligned}$ .

$\begin{aligned} & \text{length of SB} \\ =\; & (75\; \rm mi) \times \frac{\sin(40^{\circ})}{\sin(119^{\circ})} \\ \approx\; & 55.12\; \rm mi\end{aligned}$ .

In other words, the distance between the northern lighthouse and the boat is approximately $30.73\; \rm mi$ , whereas the distance between the southern lighthouse and the boat is approximately $55.12\; \rm mi$ . Hence the conclusion.