THIS IS A PRACTICE 6 Mark for Review In right triangle \( R S T \), the sum of the measures of angle \( R \) and angle \( S \) is 90 degrees. The value of \( \sin (R) \) is \( \frac{\sqrt{15}}{4} \). What is the value of \( \cos (S) \) ? (A) \( \frac{\sqrt{15}}{15} \) (B) \( \frac{\sqrt{15}}{4} \) (C) \( \frac{4 \sqrt{15}}{15} \) (D) \( \sqrt{15} \)
Added by Patricia F.
Close
Step 1
This means that \( R \) and \( S \) are complementary angles. Show more…
Show all steps
Your feedback will help us improve your experience
Erika Bustos and 64 other Algebra educators are ready to help you.
Ask a new question
Labs
Want to see this concept in action?
Explore this concept interactively to see how it behaves as you change inputs.
Key Concepts
Recommended Videos
If in a triangle $\mathrm{ABC}, \angle \mathrm{A}=90^{\circ}$, then $\frac{\mathrm{b}^{2}+\mathrm{c}^{2}}{\mathrm{~b}^{2}-\mathrm{c}^{2}} \sin (\mathrm{B}-\mathrm{C})$ is (a) 0 (b) 2 (c) $\frac{1}{2}$ (d) 1
In $\triangle A B C, C$ is a right angle and tan $A=1 .$ What is the value of $\sin A+\cos A ?$ $$\begin{array}{l}{\text { (A) } \frac{\sqrt{2}}{2}} \\ {\text { (B) } 1} \\ {\text { (C) } \sqrt{2}} \\ {\text { (D) } 2 \sqrt{2}}\end{array}$$
This exercise shows how to obtain radical expressions for $\sin 15^{\circ}$ and $\cos 15^{\circ} .$ In the figure, assume that $A B=B D=2$ GRAPH CAN'T COPY. (a) In the right triangle $B C D,$ note that $D C=1$ because $D C$ is opposite the $30^{\circ}$ angle and $B D=2$ Use the Pythagorean theorem to show that $B C=\sqrt{3}$ (b) Use the Pythagorean theorem to show that $A D=2 \sqrt{2+\sqrt{3}}$ (c) Show that the expression for $A D$ in part (b) is equal to $\sqrt{6}+\sqrt{2} .$ Hint: Two nonnegative quantities are equal if and only if their squares are equal. (d) Explain why $\angle B A D=\angle B D A$ (e) According to a theorem from geometry, an exterior angle of a triangle is equal to the sum of the two nonadjacent interior angles. Apply this to $\triangle A B D$ with exterior angle $D B C=30^{\circ},$ and show that $\angle B A D=15^{\circ}$ (f) Using the figure and the values that you have obtained for the lengths, conclude that $\sin 15^{\circ}=\frac{1}{\sqrt{6}+\sqrt{2}} \quad \cos 15^{\circ}=\frac{2+\sqrt{3}}{\sqrt{6}+\sqrt{2}}$ (g) Rationalize the denominators in part (f) to obtain $\sin 15^{\circ}=\frac{\sqrt{6}-\sqrt{2}}{4} \quad \cos 15^{\circ}=\frac{\sqrt{6}+\sqrt{2}}{4}$ (h) Use your calculator to check the results in part (g).
The Trigonometric Functions
Right-Triangle Trigonometry
Recommended Textbooks
Elementary and Intermediate Algebra
Algebra and Trigonometry
18,000,000+
Students on Numerade
Trusted by students at 8,000+ universities
Watch the video solution with this free unlock.
EMAIL
PASSWORD