MaxWong

This was answered yesterday: https://web2.0calc.com/questions/arithmetic-sequences_22

We always use the acute angle to describe the angle between two lines. That's why when we have a 120-degree angle between the lines, we calculate the angle adjacent to it and conclude that the lines form an angle of 60 degrees.

Note that the function is increasing everywhere (so when we find a solution to f(x) = k, that must be the unique solution).

When f(x) = 0, x = 3 since f(3) = 3 - 3 = 0 (solution is unique because f is increasing).

That means \(f^{-1}(0) = 3\).

For any x <= 3, f(x) = x - 3 <= 0.

Therefore, when f(x) = 6, x > 3.

Can you try to find the value of x from here? That value would be \(f^{-1}(6)\), and now you just add the results together.

It is correct since the angle could be \(180^\circ - \cos^{-1}\left(\dfrac16\right)\). Having a \(x^\circ\) angle is basically the same thing as having a \((180 - x)^\circ\) angle.

Very similar solved problem here:

https://web2.0calc.com/questions/product_4

Use sine law directly, \(\dfrac{BC}{\sin 43^\circ} = \dfrac{12\,\text{ m}}{\sin 99^\circ}\)

Please continue from here.

Using sine law directly gives \(\dfrac{AC}{\sin 58^\circ} = \dfrac{86}{\sin 84^\circ}\).

Can you take it from here? It is already one step from the answer.

\(\dfrac{7}{\sin B} = \dfrac 6{\sin 30^\circ}\\ 6\sin B = 7\sin 30^\circ\\ 6 \sin B = \dfrac72\)

Now this is a linear equation in sin B, so you can solve for sin B by dividing by 6 on both sides.

No matter B is acute or not, sin B is the same.

The number of such matches equals the number of ways that you can choose 2 left-handed person from 5 of them to play a match, which is \(\displaystyle\binom{5}{2} = 10\).

You can use cosine law to find c first.

\(c^2 = 8^2 + 11^2 - 2\cdot 8 \cdot 11 \cos 37^\circ\\ c \approx 6.666344593\)

Then, use cosine law again to find cos A.

\(8^2 = c^2 + 11^2 - 2\cdot c \cdot 11 \cos A\)

Substitute the value of c calculated above. Can you take it from here?