Question

Ques 1:
In ABC Triangle A is 90° and AC =x
If SinB=x what is sinC=?
Ques 2
-3<5-2x<7 express it with modules

Answer 1

Answer on Question #83744 – Math – Other

Question

1. In Triangle ABC the angle A is $90{}^{\circ}$ and AC =x. If SinB=x what is sinC=?

Solution

We know that $\mathrm{AC} = \mathrm{x}$, $\sin \mathrm{B} = \mathrm{x}$, $\mathrm{A} = 90{}^{\circ}$. We can use the sine theorem:

$$\frac {A C}{\sin B} = \frac {B C}{\sin A} \leftrightarrow \frac {x}{x} = \frac {B C}{\sin 9 0 {}^ {\circ}} \leftrightarrow B C = 1$$

Now, we use the Pythagorean theorem:

$$A B ^ {2} = B C ^ {2} - A C ^ {2} = 1 - x ^ {2} \leftrightarrow A B = \sqrt {1 - x ^ {2}}$$

(One must note that the same result can be obtained using the cosine theorem:

$$B C ^ {2} = A B ^ {2} + A C ^ {2} - 2 * A C * A B * \cos A \leftrightarrow 1 = A B ^ {2} + x ^ {2} - 2 * A B * A C * 0$$ $$A B ^ {2} = 1 - x ^ {2} \leftrightarrow A B = \sqrt {1 - x ^ {2}}$$

Using the definition of the sine function

$$\sin C = \frac {A B}{B C} = \frac {\sqrt {1 - x ^ {2}}}{1} = \sqrt {1 - x ^ {2}}$$

(One must note that $\sin C$ can be found using the sine theorem:

$$\frac {A B}{\sin C} = \frac {B C}{\sin A} \leftrightarrow \frac {\sqrt {1 - x ^ {2}}}{\sin C} = \frac {1}{\sin 9 0 {}^ {\circ}} \leftrightarrow \sin C = \sqrt {1 - x ^ {2}}$$

Answer: $\sin C = \sqrt{1 - x^2}$.

Question

2. -3<5-2x<7 express it with modules.

Solution

$$-3 < 5 - 2x < 7$$

Subtract 5

$$-8 < -2x < 2$$

Divide by -1

$$8 > 2x > -2$$

Subtract 3

$$5 > 2x - 3 > -5$$

Rewrite the last formula as

$$-5 < 2x - 3 < 5$$

By the definition of the absolute value it is equivalent to

$$|2x - 3| < 5$$

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