Math101 · Independent Ontario learning library
Math101learn.math101.caArea of an Oblique Triangle
An oblique triangle has no right angle. Its area can be found from two sides and their included angle: $A=\frac12ab\sin C$, where $C$ lies between sides $a$ and $b$.
The sine-area formula handles surveying, navigation, and design when altitude is not directly known. It also leads to Heron-type and law-of-sines area relationships.
Intuition and core definition
An oblique triangle has no right angle. Its area can be found from two sides and their included angle: $A=\frac12ab\sin C$, where $C$ lies between sides $a$ and $b$. Equivalent forms cycle the labels. The formula comes from height $h=b\sin C$ in the familiar $A=\frac12(\text{base})(\text{height})$.
Notation, language, and conditions
Standard notation places side $a$ opposite angle $A$, and similarly for $b,c$. The included angle is formed by the two given sides. Angles must match the calculator’s degree or radian mode. Since area is nonnegative, an ordinary triangle uses $0^\circ<C<180^\circ$ and $\sin C>0$.
Why this idea matters
The formula $A=\frac12 ab\sin C$ turns two sides and their included angle into area by extracting the perpendicular height.
A dependable method
- Sketch and label the two known sides and their included angle.
- Verify that the angle is between those sides.
- Choose the matching form $\frac12ab\sin C$.
- Set calculator angle mode, substitute, and preserve guard digits.
- Report square units and check against $ab/2$, the maximum for those sides.
Worked example
Representations and interpretation
Drop an altitude from the endpoint of one known side. In the resulting right triangle, height equals one side times the sine of the included angle; the other known side serves as base. This construction explains both the formula and its angle condition.
Reasoning about variations
Angles $C$ and $180^\circ-C$ have the same sine, so the same two side lengths can produce equal areas for an acute or obtuse included angle. Their shapes differ even though base-height product matches.
Common mistakes
How to check your work
- Ensure area is positive and no greater than $ab/2$.
- Compute the corresponding altitude and use base-height area.
- Estimate sine from the angle’s quadrant and benchmark values.
Practice
- Find the area for sides $10$ cm and $14$ cm with included angle $30^\circ$.
- Which angle belongs in $\frac12ab\sin C$?
- What is the greatest possible area for fixed sides $8$ and $11$?
Answers and brief solutions
Show answers
- $35$ cm$^2$ $\frac12(10)(14)\sin30^\circ=70(1/2)=35$.
- The angle included between sides $a$ and $b$ $C$ is opposite $c$ and formed by sides $a,b$.
- $44$ square units Maximum sine is $1$ at a $90^\circ$ included angle.
Synthesis and transfer
A land parcel measured along two boundaries and their included bearing can be split conceptually into base and height; an obtuse angle still gives positive area because its sine is positive.
Suppose two property boundaries from one survey marker measure $80$ m and $55$ m with included angle $125^\circ$. The area is $\frac12(80)(55)\sin125^\circ$, about $1802$ m$^2$. Dropping a perpendicular from the third vertex explains the sine: the height relative to the $80$ m base is $55\sin125^\circ$. Using the supplementary angle $55^\circ$ gives the same area because the two sines are equal, consistent with mirrored triangles sharing base and height. The included-angle requirement matters; a nonincluded angle paired with the same two lengths would not determine this calculation directly. Square units and an estimate below $80\cdot55/2$ check plausibility.
Related topics
Teaching and accessibility note
Try it yourself
Hints are part of learning. Open one whenever it makes the next step feel possible.
Find the area for sides $10$ cm and $14$ cm with included angle $30^\circ$.
- $\frac12(10)(14)\sin30^\circ=70(1/2)=35$.
End of lesson
Nice work making it this far.
Understanding grows through return visits. Save this lesson, try the practice, or continue when you are ready.
