Question

find the exact value of \csc b in simplest form.
triangle with vertices d, c (right angle), b; side dc=20, bc=21, bd=29
answer attempt 1 out of 2
\csc b = input box submit answer

Explanation:

Step1: Recall the definition of cosecant

The cosecant of an angle in a right triangle is the reciprocal of the sine of that angle. For angle \( B \) in right triangle \( BCD \) (right - angled at \( C \)), \(\csc B=\frac{1}{\sin B}\), and \(\sin B = \frac{\text{opposite side to }B}{\text{hypotenuse}}\).

Step2: Identify the sides relative to angle \( B \)

In right triangle \( BCD \), the side opposite to angle \( B \) is \( CD = 20 \), and the hypotenuse is \( BD=29 \). Wait, no, wait. Wait, in right - angled triangle at \( C \), the sides: \( BC = 21 \), \( CD = 20 \), \( BD = 29 \) (since \( 21^{2}+20^{2}=441 + 400=841=29^{2}\), so by Pythagoras theorem, \( BD \) is the hypotenuse). The side opposite to angle \( B \) is \( CD = 20 \)? No, wait, angle \( B \) is at vertex \( B \), so the sides: adjacent side to \( B \) is \( BC = 21 \), opposite side to \( B \) is \( CD = 20 \), and hypotenuse is \( BD = 29 \). But \(\sin B=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{CD}{BD}=\frac{20}{29}\)? Wait, no, wait. Wait, in triangle \( BCD \), right - angled at \( C \), angle at \( B \): the opposite side to \( B \) is \( CD \), adjacent is \( BC \), hypotenuse is \( BD \). But \(\csc B=\frac{1}{\sin B}=\frac{\text{hypotenuse}}{\text{opposite side to }B}\). Wait, no, \(\sin B=\frac{\text{opposite}}{ \text{hypotenuse}}\), so \(\csc B=\frac{\text{hypotenuse}}{\text{opposite}}\). Wait, let's re - check. The definition of sine in a right triangle: \(\sin\theta=\frac{\text{length of opposite side}}{\text{length of hypotenuse}}\), so \(\csc\theta=\frac{1}{\sin\theta}=\frac{\text{length of hypotenuse}}{\text{length of opposite side}}\).

In triangle \( BCD \), right - angled at \( C \), for angle \( B \):
- Opposite side: \( CD = 20 \)
- Hypotenuse: \( BD = 29 \)
- Adjacent side: \( BC = 21 \)

Wait, no, that's a mistake. Wait, angle \( B \): the sides: the side opposite to \( B \) is \( CD \), the side adjacent to \( B \) is \( BC \), and the hypotenuse is \( BD \). But let's confirm with Pythagoras: \( BC^{2}+CD^{2}=21^{2}+20^{2}=441 + 400 = 841=29^{2}=BD^{2}\), so that's correct.

But wait, \(\sin B=\frac{CD}{BD}=\frac{20}{29}\), so \(\csc B=\frac{BD}{CD}=\frac{29}{20}\)? Wait, no, that can't be. Wait, no, I think I mixed up the opposite side. Wait, angle \( B \): the vertex is \( B \), so the sides: from \( B \), the two legs are \( BC \) and \( BD \)? No, no, the triangle is \( B - C - D \), with right angle at \( C \). So the sides are \( BC \), \( CD \), and \( BD \). So angle at \( B \): the sides:
- Opposite side: \( CD \) (because it's across from \( B \))
- Adjacent side: \( BC \) (because it's one of the legs forming angle \( B \))
- Hypotenuse: \( BD \) (the side opposite the right angle)

Wait, but let's think again. The sine of angle \( B \) is equal to the length of the side opposite angle \( B \) divided by the length of the hypotenuse. The side opposite angle \( B \) is \( CD = 20 \), hypotenuse is \( BD = 29 \). So \(\sin B=\frac{20}{29}\), then \(\csc B=\frac{1}{\sin B}=\frac{29}{20}\). Wait, but let's check with the other leg. Wait, maybe I got the opposite side wrong. Wait, angle \( B \): the angle at \( B \), so the sides: the side opposite to \( B \) is \( CD \), and the side adjacent is \( BC \). Wait, but if we consider angle \( B \), the triangle is \( \triangle BCD \), right - angled at \( C \). So \( \angle C = 90^{\circ} \), so \( \sin B=\frac{CD}{BD}\), \( \cos B=\frac{BC}{BD}\). So \( \csc B=\frac{BD}{CD}=\frac{29}{20}\)? Wait, but let's check the lengths again. \( BC = 21 \), \( CD = 20 \), \( BD = 29 \). So yes, \( 21^{2}+20^{2}=441 + 400 = 841=29^{2}\), so that's a right triangle. So \(\csc B=\frac{\text{hypotenuse}}{\text{opposite side to }B}=\frac{BD}{CD}=\frac{29}{20}\).

Wait, no, wait a second. I think I made a mistake in identifying the opposite side. Let's label the triangle properly. Let's say \( B \) is a vertex, \( C \) is the right - angled vertex, so \( BC \) and \( CD \) are the legs, \( BD \) is the hypotenuse. So angle at \( B \): the sides: the side opposite angle \( B \) is \( CD \), the side adjacent is \( BC \), and hypotenuse is \( BD \). So \(\sin B=\frac{CD}{BD}=\frac{20}{29}\), so \(\csc B=\frac{BD}{CD}=\frac{29}{20}\).

Answer:

\(\frac{29}{20}\)