Question

find the sine, cosine, and tangent of ∠d.

(there is a right triangle with right angle at b. side cb is 4, side cd is 2√15, vertices are c, b, d with d at the top, b at the right angle, c at the left.)

write your answer in simplified, rationalized form. do not round.

sin(d) =
cos(d) =
tan(d) =

Explanation:

Step1: Find the length of BD

In right triangle \(BCD\), by Pythagorean theorem \(CD^{2}=BC^{2}+BD^{2}\). Given \(CD = 2\sqrt{15}\), \(BC = 4\). So \(BD=\sqrt{CD^{2}-BC^{2}}=\sqrt{(2\sqrt{15})^{2}-4^{2}}=\sqrt{60 - 16}=\sqrt{44}=2\sqrt{11}\)? Wait, no, wait, \( (2\sqrt{15})^2=4\times15 = 60\), \(4^2 = 16\), \(60-16 = 44\)? Wait, no, maybe I made a mistake. Wait, \(CD\) is the hypotenuse? Wait, the right angle is at \(B\), so \(BC\) and \(BD\) are legs, \(CD\) is hypotenuse. So \(CD^{2}=BC^{2}+BD^{2}\), so \(BD=\sqrt{CD^{2}-BC^{2}}=\sqrt{(2\sqrt{15})^{2}-4^{2}}=\sqrt{60 - 16}=\sqrt{44}=2\sqrt{11}\)? Wait, no, wait, maybe I mixed up the sides. Wait, angle \(D\): in angle \(D\), the opposite side is \(BC\), adjacent side is \(BD\), hypotenuse is \(CD\). Wait, let's re - identify: in right triangle \(BCD\), right - angled at \(B\). So for \(\angle D\):
- Opposite side (\(opp\)): \(BC = 4\)
- Adjacent side (\(adj\)): \(BD\)
- Hypotenuse (\(hyp\)): \(CD=2\sqrt{15}\)

First, find \(BD\) using Pythagoras: \(BD=\sqrt{CD^{2}-BC^{2}}=\sqrt{(2\sqrt{15})^{2}-4^{2}}=\sqrt{60 - 16}=\sqrt{44}=2\sqrt{11}\)? Wait, no, \(60 - 16=44\), \(\sqrt{44}=2\sqrt{11}\). Wait, but maybe I made a mistake. Wait, let's check again: \( (2\sqrt{15})^2=4\times15 = 60\), \(4^2 = 16\), \(60 - 16 = 44\), so \(BD=\sqrt{44}=2\sqrt{11}\).

Step2: Calculate \(\sin(D)\)

The formula for sine of an angle in a right triangle is \(\sin(\theta)=\frac{\text{opposite}}{\text{hypotenuse}}\). For \(\angle D\), opposite side is \(BC = 4\), hypotenuse is \(CD = 2\sqrt{15}\). So \(\sin(D)=\frac{BC}{CD}=\frac{4}{2\sqrt{15}}=\frac{2}{\sqrt{15}}=\frac{2\sqrt{15}}{15}\) (rationalizing the denominator: multiply numerator and denominator by \(\sqrt{15}\), \(\frac{2\times\sqrt{15}}{\sqrt{15}\times\sqrt{15}}=\frac{2\sqrt{15}}{15}\))

Step3: Calculate \(\cos(D)\)

The formula for cosine of an angle in a right triangle is \(\cos(\theta)=\frac{\text{adjacent}}{\text{hypotenuse}}\). For \(\angle D\), adjacent side is \(BD = 2\sqrt{11}\)? Wait, no, wait, I think I messed up the adjacent and opposite sides. Wait, in \(\angle D\), the sides:
- Opposite: \(BC\) (since it's opposite to \(\angle D\))
- Adjacent: \(BD\) (since it's adjacent to \(\angle D\) and part of the right angle)
- Hypotenuse: \(CD\)

Wait, but let's re - derive. Let's use the definitions:

\(\sin(D)=\frac{\text{opposite to }D}{\text{hypotenuse}}=\frac{BC}{CD}\)

\(\cos(D)=\frac{\text{adjacent to }D}{\text{hypotenuse}}=\frac{BD}{CD}\)

\(\tan(D)=\frac{\text{opposite to }D}{\text{adjacent to }D}=\frac{BC}{BD}\)

We have \(BC = 4\), \(CD = 2\sqrt{15}\), \(BD=\sqrt{CD^{2}-BC^{2}}=\sqrt{60 - 16}=\sqrt{44}=2\sqrt{11}\)? Wait, no, \(60-16 = 44\), \(\sqrt{44}=2\sqrt{11}\). Wait, but let's check the calculation of \(BD\) again. Wait, \(CD\) is the hypotenuse, so \(CD^{2}=BC^{2}+BD^{2}\), so \(BD^{2}=CD^{2}-BC^{2}=(2\sqrt{15})^{2}-4^{2}=60 - 16 = 44\), so \(BD = \sqrt{44}=2\sqrt{11}\).

Now, \(\cos(D)=\frac{BD}{CD}=\frac{2\sqrt{11}}{2\sqrt{15}}=\frac{\sqrt{11}}{\sqrt{15}}=\frac{\sqrt{165}}{15}\) (rationalizing: multiply numerator and denominator by \(\sqrt{15}\), \(\frac{\sqrt{11}\times\sqrt{15}}{\sqrt{15}\times\sqrt{15}}=\frac{\sqrt{165}}{15}\))

Step4: Calculate \(\tan(D)\)

\(\tan(D)=\frac{\text{opposite}}{\text{adjacent}}=\frac{BC}{BD}=\frac{4}{2\sqrt{11}}=\frac{2}{\sqrt{11}}=\frac{2\sqrt{11}}{11}\) (rationalizing: multiply numerator and denominator by \(\sqrt{11}\))

Wait, but wait, maybe I made a mistake in identifying the sides. Wait, let's re - examine the triangle. The triangle is labeled \(C\), \(B\), \(D\) with right angle at \(B\). So the vertices are \(C\)---\(B\)---\(D\) with \(CB = 4\), \(CD = 2\sqrt{15}\), right angle at \(B\). So:

- For \(\angle D\):
- The side opposite to \(\angle D\) is \(CB = 4\)
- The side adjacent to \(\angle D\) is \(BD\)
- The hypotenuse is \(CD=2\sqrt{15}\)

So first, find \(BD\):

\(BD=\sqrt{CD^{2}-CB^{2}}=\sqrt{(2\sqrt{15})^{2}-4^{2}}=\sqrt{60 - 16}=\sqrt{44}=2\sqrt{11}\)

Now,

- \(\sin(D)=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{CB}{CD}=\frac{4}{2\sqrt{15}}=\frac{2}{\sqrt{15}}=\frac{2\sqrt{15}}{15}\) (rationalizing the denominator: multiply numerator and denominator by \(\sqrt{15}\))
- \(\cos(D)=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{BD}{CD}=\frac{2\sqrt{11}}{2\sqrt{15}}=\frac{\sqrt{11}}{\sqrt{15}}=\frac{\sqrt{11}\times\sqrt{15}}{\sqrt{15}\times\sqrt{15}}=\frac{\sqrt{165}}{15}\)
- \(\tan(D)=\frac{\text{opposite}}{\text{adjacent}}=\frac{CB}{BD}=\frac{4}{2\sqrt{11}}=\frac{2}{\sqrt{11}}=\frac{2\sqrt{11}}{11}\) (rationalizing the denominator: multiply numerator and denominator by \(\sqrt{11}\))

Answer:

\(\sin(D)=\frac{2\sqrt{15}}{15}\)

\(\cos(D)=\frac{\sqrt{165}}{15}\)

\(\tan(D)=\frac{2\sqrt{11}}{11}\)