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Equilateral Triangle

full face gasket
  • A+B+C=180^\circ
  • Find Area \large%20\ A=\frac sqrt{3}{4}(a^2)

Oblique 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
full face gasket
full face gasket
  • Find p
    • given a b c: \large%20\ p= a+b+c
  • Find s
    • given a b c: \large%20\ s= (a+b+c) \div 2
  • Find d
    • given a b c s: \large%20\ d= (b^2+c^2-a^2) \div 2b
  • Find e
    • given a b c s: \large%20\ e= (a^2+b^2-c^2) \div 2b
  • 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)

Right Triangle


  • 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.
full face gasket
  • 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)
  • Find Area
    • given a b: \large%20\ Area= ab \div2

Square Pyramid

  • 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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