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katrin2010 [14]
3 years ago
5

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.

4 0
3 years ago
A) -3 · x = -21<br><br> b) -7x + 16 = 2x - 20<br> help pls
NeTakaya

Answer:4

Step-by-step explanation:

B) -7x-2x= -20-16

    -9x= -36

      X= -36 /-9

      X= 4

7 0
3 years ago
7176 rounded to 1 significant figure
kati45 [8]

Answer:

7000

Step-by-step explanation:

8 0
3 years ago
Pat made a total of 48 pottery plates and cups. if she made twice as many plates as cups, how many plates did she make?
ddd [48]
If C stands for cups, solve for C

48 - 2c = 0
    l
subtract 48 from each side.
    l
2c=-48
    l
divide 48 by 2 --->  24.



3 0
3 years ago
Read 2 more answers
There is a parallelogram ABCD with diagonals AC and BD. The diagonals AC and BD intersects each other at point E. Side AB is con
grandymaker [24]

Answer:

SAS theorem

Step-by-step explanation:

Given

\square ABCD

\[ \lvert \[ \lvert AB =\[ \lvert \[ \lvert CD

\angle BAC = \angle  DCA

Required

Which theorem shows △ABE ≅ △CDE.

From the question, we understand that:

AC and BD intersects at E.

This implies that:

\[ \lvert \[ \lvert AE = \[ \lvert \[ \lvert EC

and

\[ \lvert \[ \lvert BE = \[ \lvert \[ \lvert ED

So, the congruent sides and angles of △ABE and △CDE are:

\[ \lvert \[ \lvert AB =\[ \lvert \[ \lvert CD ---- S

\angle BAC = \angle  DCA ---- A

\[ \lvert \[ \lvert BE = \[ \lvert \[ \lvert ED or \[ \lvert \[ \lvert AE = \[ \lvert \[ \lvert EC  --- S

<em>Hence, the theorem that compares both triangles is the SAS theorem</em>

4 0
2 years ago
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