amanf
2020-12-24
Answered

Solve
$\mathrm{cos}30\cdot \frac{\mathrm{sin}(-50)}{\mathrm{tan}70}$

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Derrick

Answered 2020-12-25
Author has **94** answers

so the expression becomes

asked 2021-06-16

Airline passengers arrive randomly and independently at the passenger-screening facility at a major international airport. The mean arrival rate is 11 passengers per minute.

asked 2020-10-18

If $\mathrm{sin}x+\mathrm{sin}y=a{\textstyle \phantom{\rule{1em}{0ex}}}\text{and}{\textstyle \phantom{\rule{1em}{0ex}}}\mathrm{cos}x+\mathrm{cos}y=b$ then find $\mathrm{tan}(x-\frac{y}{2})$

asked 2021-12-30

We have:

$\{\begin{array}{c}2x=y\mathrm{tan}\theta +\mathrm{sin}\theta \\ 2y=x\mathrm{cot}\theta +\mathrm{cos}\theta \end{array}$

And want to prove${x}^{2}+{y}^{2}=1$

My works:

I multiplied first equation by$\mathrm{cos}\theta$ and second one by $\mathrm{sin}\theta$ and get:

$\{\begin{array}{c}2x\mathrm{cos}\theta =y\mathrm{sin}\theta +\mathrm{sin}\theta \mathrm{cos}\theta \\ 2y\mathrm{sin}\theta =x\mathrm{cos}\theta +\mathrm{sin}\theta \mathrm{cos}\theta \end{array}$

By extracting$\mathrm{sin}\theta \mathrm{cos}\theta$ we get:

$2x\mathrm{cos}\theta -y\mathrm{sin}\theta =2y\mathrm{sin}\theta -x\mathrm{cos}\theta$

$x\mathrm{cos}\theta =y\mathrm{sin}\theta$

And want to prove

My works:

I multiplied first equation by

By extracting

asked 2021-05-28

28.116∘30′

asked 2021-11-06

use the given function value(s), and trigonometric identities (including the cofunction identities), to find the indicated trigonometric functions.

$\mathrm{cos}\theta =\frac{1}{3}$

$\text{(a)}\mathrm{sin}\theta \text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{(b)}\mathrm{tan}\theta$

$\text{(c)}\mathrm{sec}\theta \text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{}\text{(d)}\mathrm{csc}(90-\theta )$

asked 2022-07-09

I have three random points, O, A, B, with these I can get angles $\alpha $ and $\beta $

How can I get the angle C, or better, directional vector of c which will be evenly in the middle, between the other two angles. Keep in mind that the angles might be of any value.

This old answer almost gets me were I need to be, but how can I change the final formula presented in the answer $g=\text{arctan}\phantom{\rule{thinmathspace}{0ex}}(2\mathrm{tan}r)$ to work with non-right triangles? to work with non-right triangles?

I have found lots of results searching for this, but they don't seem to work in my case.

It looks like a good solution, but I don't know enough about trig to determine the part to change in the final formula to make it not right triangle dependant

How can I get the angle C, or better, directional vector of c which will be evenly in the middle, between the other two angles. Keep in mind that the angles might be of any value.

This old answer almost gets me were I need to be, but how can I change the final formula presented in the answer $g=\text{arctan}\phantom{\rule{thinmathspace}{0ex}}(2\mathrm{tan}r)$ to work with non-right triangles? to work with non-right triangles?

I have found lots of results searching for this, but they don't seem to work in my case.

It looks like a good solution, but I don't know enough about trig to determine the part to change in the final formula to make it not right triangle dependant

asked 2021-11-21

Consider the non-right triangle below

Suppose that$m\mathrm{\angle}BCA={69}^{\circ}$ , and that $x=32cm$ and $y=49cm$ . What is the degree measure of $\mathrm{\angle}ABC?$

Suppose that