Question

express cos b as a fraction in simplest terms.
triangle with right angle at c, dc = √51, db = 12, vertices d, c, b

Explanation:

Step1: Identify right triangle parts

Right triangle \( \triangle BCD \), right-angled at \( C \). Hypotenuse \( BD = 12 \), adjacent side to \( \angle B \) is \( BC \), opposite is \( CD = \sqrt{51} \).

Step2: Calculate \( BC \) using Pythagoras

\( BC^2 + CD^2 = BD^2 \)
\( BC^2 + (\sqrt{51})^2 = 12^2 \)
\( BC^2 + 51 = 144 \)
\( BC^2 = 144 - 51 = 93 \)
\( BC = \sqrt{93} \)

Step3: Compute \( \cos B \)

\( \cos B = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{BC}{BD} = \frac{\sqrt{93}}{12} \)

Answer:

\( \frac{\sqrt{93}}{12} \)