3 Sides

(Redirected from Oblique Triangle)

Equilateral Triangle

An oblique triangle is any triangle that is not a Right Triangle.
An acute triangle is when all three angles of the triangle are less than right angles.
An obtuse triangle is when one of the three angles is greater than a right angle.
For "s" and "p" , see semiperimeter

Find h
- given c A: $\large%20\ h= c*sinA$
- given a C: $\large%20\ h= a*sinC$

Find Area
- given a b c: $\large%20\ Area=\frac {1}{2}(bh)$
- given a b c: $\large%20\ Area= sqrt[s(s-a)(s-<img src='http://www.piping-designer.com/board/public/style_emoticons/default/cool.png' border='0' alt='user posted image' />(s-c)]$
- given C a b: $\large%20\ Area= \frac{1}{2} ab* \sin c$

Find A
- given a b c s: $\large%20\ \sin \frac{1}{2} A= sqrt[(s-<img src='http://www.piping-designer.com/board/public/style_emoticons/default/cool.png' border='0' alt='user posted image' />(s-c) \div bc]$
- given a b c s: $\large%20\ \cos \frac{1}{2} A= sqrt[s(s-a) \div bc]$
- given a b c s: $\large%20\ \tan \frac{1}{2} A= sqrt[(s-<img src='http://www.piping-designer.com/board/public/style_emoticons/default/cool.png' border='0' alt='user posted image' />(s-c) \div s(s-a)]$
- given c h: $\large%20\ sin A= \frac {h}{c}$
- given B a b: $\large%20\ sin A= a* sin B\div b$
- given B a c: $\large%20\ A= \frac{1}{2}(A+C)+\frac{1}{2}(A-C}$
- given C a b: $\large%20\ A= \frac{1}{2}(A+<img src='http://www.piping-designer.com/board/public/style_emoticons/default/cool.png' border='0' alt='user posted image' />+\frac{1}{2}(A-B}$
- given C a c: $\large%20\ sin A= a* sin C \div c$

Find B
- given a b c s: $\large%20\ \sin \frac{1}{2} B= sqrt[(s-a)(s-c) \div ac]$
- given a b c s: $\large%20\ \cos \frac{1}{2} B= sqrt[s(s-<img src='http://www.piping-designer.com/board/public/style_emoticons/default/cool.png' border='0' alt='user posted image' /> \div ac]$
- given a b c s: $\large%20\ \tan \frac{1}{2} B= sqrt[(s-a)(s-c) \div s(s-<img src='http://www.piping-designer.com/board/public/style_emoticons/default/cool.png' border='0' alt='user posted image' />]$
- given a h: $\large%20\ sin B= \frac {h}{a}$
- given A a b: $\large%20\ sin B= b* sin A \div a$
- given A b c: $\large%20\ B= \frac{1}{2}(B+C)+\frac{1}{2}(B-C}$
- given C a b: $\large%20\ B= \frac{1}{2}(A+<img src='http://www.piping-designer.com/board/public/style_emoticons/default/cool.png' border='0' alt='user posted image' />-\frac{1}{2}(A-B}$
- given C a c: $\large%20\ sin B= b* sin C \div c$

Find C
- given a b c s: $\large%20\ \sin \frac{1}{2} C= sqrt[(s-a)(s-<img src='http://www.piping-designer.com/board/public/style_emoticons/default/cool.png' border='0' alt='user posted image' /> \div ab]$
- given a b c s: $\large%20\ \cos \frac{1}{2} C= sqrt[s(s-c) \div ab]$
- given a b c s: $\large%20\ \tan \frac{1}{2} C= sqrt[(s-a)(s-<img src='http://www.piping-designer.com/board/public/style_emoticons/default/cool.png' border='0' alt='user posted image' /> \div s(s-c)]$
- given A a c: $\large%20\ sin C= c* sin A \div a$
- given A b c: $\large%20\ C= \frac{1}{2}(B+C)-\frac{1}{2}(B-C}$
- given B a c: $\large%20\ C= \frac{1}{2}(A+C)-\frac{1}{2}(A-C}$
- given B b c: $\large%20\ sin C= c* sin B \div c$

Find a
- given A B b: $\large%20\ a= b* \sin A \div \sin B$
- given A B c: $\large%20\ a= c* \sin A \div \sin (A+<img src='http://www.piping-designer.com/board/public/style_emoticons/default/cool.png' border='0' alt='user posted image' />$
- given A C b: $\large%20\ a= b* \sin A \div \sin (A+C)$
- given A C c: $\large%20\ a= c* \sin A \div \sin C$
- given B C b: $\large%20\ a= b* \sin (A+C) \div \sin B$
- given B C c: $\large%20\ a= c* \sin (A+C) \div \sin C$
- given A b c: $\large%20\ a= sqrt(b^2+c^2-2bc * \cos A)$

Find b
- given A B a: $\large%20\ b= a* \sin B \div \sin A$
- given A B c: $\large%20\ b= c* \sin B \div \sin (A+<img src='http://www.piping-designer.com/board/public/style_emoticons/default/cool.png' border='0' alt='user posted image' />$
- given A C a: $\large%20\ b= a* \sin (A+C) \div \sin A$
- given A C c: $\large%20\ b= c* \sin (A+C) \div \sin C$
- given B C a: $\large%20\ b= a* \sin B \div \sin (B+C)$
- given B C c: $\large%20\ b= c* \sin B \div \sin C$
- given B a c: $\large%20\ b= sqrt(a^2+c^2-2ac * \cos <img src='http://www.piping-designer.com/board/public/style_emoticons/default/cool.png' border='0' alt='user posted image' />$

Find c
- given A B a: $\large%20\ c= a* \sin (A+<img src='http://www.piping-designer.com/board/public/style_emoticons/default/cool.png' border='0' alt='user posted image' /> \div \sin A$
- given A B b: $\large%20\ c= b* \sin (A+<img src='http://www.piping-designer.com/board/public/style_emoticons/default/cool.png' border='0' alt='user posted image' /> \div \sin B$
- given A C a: $\large%20\ c= a* \sin C \div \sin A$
- given A C b: $\large%20\ c= b* \sin C \div \sin (A+C)$
- given B C a: $\large%20\ c= a* \sin C \div \sin (B+C)$
- given B C b: $\large%20\ c= b* \sin C \div \sin B$
- given C a b: $\large%20\ c= sqrt(a^2+b^2-2ab * \cos C)$

One side of a right triangle is 90.
The hypotenuse of a right triangle is the longest side or the side opposite the right angle.

Find A
- given a c: $\large%20\ sin A= a \div c$
- given b c: $\large%20\ cos A= b \div c$
- given a b: $\large%20\ tan A= a \div b$

Find B
- given a c: $\large%20\ sin B= a \div c$
- given b c: $\large%20\ cos B= b \div c$
- given a b: $\large%20\ tan B= b \div a$

Find a
- given A c: $\large%20\ a= c*sin A$
- given A b: $\large%20\ a= b*tan A$

Find b
- given A c: $\large%20\ b= c*cos A$
- given A a: $\large%20\ b= a \div tan A$

Find c
- given A a: $\large%20\ c= a \div sin A$
- given A b: $\large%20\ c= b \div cos A$
- given a b: $\large%20\ c= sqrt (a^2+b^2)$

Volume $\large%20\ V=\frac{1}{3}\ abh$
Slant height $\large%20\ s= sqrt (r^2+h^2)$
Base perimeter $\large%20\ p= 2(a+<img src='http://www.piping-designer.com/board/public/style_emoticons/default/cool.png' border='0' alt='user posted image' />$
Laterial surface area $\large%20\ L= \frac{1}{2}ps$
Surface area $\large%20\ A= 2as+(ab)^2$

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