Math Forum/Help
College Math
Adding and subtracting up to 10
Comparing numbers up to 10
Addition and Subtraction within 20
Adding and Subtracting up to 100
Addition and Subtraction within 1000
Multiplication up to 5
Addition, Multiplication, Division
Adding and Subtracting
Divisibility by 2, 3, 4, 5, 9
Area of Squares and Rectangles
Least Common Multiple
Fraction Multiplication and Division
Mixed Numbers
Signed Numbers
The Coordinate Plane
Symplifying Expressions
Polynomial Vocabulary
Polynomial Expressions
Linear Functions
Systems of equations
Parametric Linear Equations
Vieta's Formulas
Geometric Progression
Reciprocal Equations
Logarithmic Expressions
Logarithmic Equations
Extremal value problems
Numbers Classification
Intercept Theorem
Law of Sines
Limits of Functions
Properties of Triangles
Law of Sines: Problems with Solutions
What is the length of $\overline{AB}$?
$\sin 30^{\circ }=\frac{\overline{BC}}{\overline{AB}}$
$\overline{AB}=\frac{\overline{BC}}{\sin 30^{\circ }}=\frac{2}{\frac{1}{2}}=4$
$\overline{AB}=4$
Calculate $\gamma ,a,c$.
$\gamma =30^{\circ }\qquad a=10\qquad c=10$
$\gamma =30^{\circ }\qquad a=13\qquad c=8$
$\gamma =30^{\circ }\qquad a=13.44\qquad c=8.76$
Answer: $\gamma =30^{\circ }\qquad a=13.44\qquad c=8.76$
According to the sine rule
$\frac{\sin A}{BC}=\frac{\sin B}{AC}=\frac{\sin C}{AB}$
then $\frac{\sin A}{BC}=\frac{\sin B}{AC}$
$BC=AC\frac{\sin A}{\sin B}=6\frac{\sin 130^{\circ }}{\sin 20^{\circ }}=6\cdot \frac{\sin \frac{5}{18}\pi }{\sin \frac{1}{9}\pi }=6\times 2.24= 13.44$
So $a=BC=13.44$
Now $m\angle A+m\angle B+m\angle C=180^{\circ }$
$m\angle C=180^{\circ }-20^{\circ }-130^{\circ }=30^{\circ }$
So $\gamma =30^{\circ }$
According to the sine rule $\frac{\sin C}{AB}=\frac{\sin B}{AC}$ then
$AB=AC\frac{\sin C}{\sin B}=6\frac{\sin 30^{\circ }}{\sin 20^{\circ }}=6 \times 1.46= 8.76$ then
$c=8.76$
Given a triangle ABC, in which [tex]AB=\sqrt{2}[/tex] and [tex]\angle ACB=45^{\circ}[/tex], determine the length of the circumradius (radius of the circumscribed circle).
By the law of sines, we have [tex]\frac{AB}{\sin \angle ACB}=2R[/tex], so [tex]R=\frac{AB}{2\sin\angle ACB}=\frac{\sqrt{2}}{2 \frac{\sqrt{2}}{2}}=1[/tex]
Given a triangle ABC, in which [tex]AB=2\sqrt{3}[/tex] and [tex]\angle ACB=60^{\circ}[/tex], determine the length of the circumradius (radius of the circumscribed circle).
By the law of sines, we have [tex]\frac{AB}{\sin \angle ACB}=2R[/tex], so [tex]R=\frac{AB}{2\sin\angle ACB}=\frac{2\sqrt{3}}{2 \frac{\sqrt{3}}{2}}=2[/tex]
Given a triangle ABC, in which [tex]AB=5[/tex] and [tex]\angle ACB=30^{\circ}[/tex], determine the length of the circumradius (radius of the circumscribed circle).
By the law of sines, we have [tex]\frac{AB}{\sin \angle ACB}=2R[/tex], so [tex]R=\frac{AB}{2\sin\angle ACB}=\frac{5}{2 \frac{1}{2}}=5[/tex]
Submit a problem on this page.
Problem text:
Your name(if you would like to be published):
E-mail(you will be notified when the problem is published)
Notes: use [tex][/tex] (as in the forum if you would like to use latex).
Wrong:
Unsolved problems:
Follow us on Twitter Facebook