Triangle Area with Sine Advanced. This lesson goes beyond core Geometry. You can skip it.

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Learning goals

Read sine and cosine from the coordinates of a point on the unit semicircle
Explain why supplementary angles give equal heights and opposite horizontal projections
Derive the triangle area formula for acute, right, and obtuse included angles
Find exact and approximate areas from two sides and their included angle

From a right triangle to a semicircle

In [Sine, Cosine, and Tangent](/geometry/right-triangles/right-triangle-trigonometry/), you found sine and cosine by dividing right-triangle leg lengths by the hypotenuse. For a hypotenuse of length $1$ , the ratios are just the leg lengths. Coordinates let us use the same idea when the angle opens past $90^\circ$ .

Draw the upper half of the circle with radius $1$ centered at the origin. Start on the positive $x$ -axis and turn counterclockwise through $\theta$ degrees, from $0^\circ$ to $180^\circ$ . Let $P=(x,y)$ be the point where the ray meets the unit semicircle.

\begin{aligned} \cos\theta&=x,\\ \sin\theta&=y. \end{aligned}

For an acute angle, dropping a perpendicular from $P$ to the horizontal axis gives a right triangle. Its horizontal leg is adjacent to the angle, its vertical leg is opposite, and its hypotenuse has length $1$ . Thus the coordinate definitions agree with the ratios you already know.

The $x$ -coordinate is the signed horizontal projection of the radius. A point to the right of the vertical axis gives a positive projection. A point to the left gives a negative one. Its distance from the vertical axis is $|x|$ . The height above the horizontal axis is $y$ , which stays nonnegative on the upper semicircle.

Reflect across the vertical axis. The height stays fixed. The projections have the same magnitude p and opposite signs.

At the three axis points, read the coordinates directly.

Turn	Point on the semicircle
$0^\circ$	$(1,0)$
$90^\circ$	$(0,1)$
$180^\circ$	$(-1,0)$

Check your understanding

What are the sine and cosine of $180^\circ$ , in that order?

0

and

1

-1

and

0

0

and

-1

1

and

0

What a reflection preserves

Two angles are supplementary when their measures add to $180^\circ$ . A ray at $30^\circ$ makes a $30^\circ$ angle with the positive axis. Reflect it in the vertical axis and the new ray makes a $30^\circ$ angle with the negative axis. Its turn from the positive axis is now $150^\circ$ . The two endpoints on the semicircle have the same height, and their horizontal coordinates have opposite signs.

Sine and cosine of a supplementary angle#

The same reflection sends the radius at any angle $\theta$ to the radius at $180^\circ-\theta$ . The two turns fit a half-turn because the reflected ray makes the same angle with the negative axis that the original ray makes with the positive axis. Reflection preserves $y$ and replaces $x$ by $-x$ . Reading the coordinates gives

\begin{aligned} \sin(180^\circ-\theta)&=\sin\theta,\\ \cos(180^\circ-\theta)&=-\cos\theta. \end{aligned}

This includes the two ends of the semicircle and the top, where the horizontal coordinate is zero.

Worked example 1 Read an obtuse angle from a familiar acute angle

Problem. Find the exact sine and cosine of $120^\circ$ using the exact values at $60^\circ$ .

Solution. The supplement of $120^\circ$ is $60^\circ$ . Keep the sine and reverse the sign of the cosine.

\begin{aligned} \sin120^\circ&=\sin60^\circ,\\ \sin120^\circ&=\frac{\sqrt3}{2}. \end{aligned}

\begin{aligned} \cos120^\circ&=-\cos60^\circ,\\ \cos120^\circ&=-\frac12. \end{aligned}

Check your understanding

The unit radius at $74^\circ$ ends at $(u,v)$ . Where does the unit radius at $106^\circ$ end?

(u,-v)

(-u,v)

(-u,-v)

(v,u)

Scale the projections

Place one end of a segment of length $r$ at the origin. Let it make an angle $\theta$ above the positive horizontal axis. Its endpoint lies $r$ times as far along the ray as the unit point, so a dilation about the origin multiplies both coordinates by $r$ .

(x,y)=(r\cos\theta,\ r\sin\theta).

The projection $r\cos\theta$ records left or right. The height $r\sin\theta$ records the perpendicular distance to the horizontal line.

Worked example 2 Locate the tip of a tilted sign

Problem. A sign arm is $130$ cm long. Its hinge is at the origin, and the arm points up and left. The unit point on its ray has coordinates $\left(-\tfrac{16}{65},\tfrac{63}{65}\right)$ . Find the signed horizontal projection and the height of the tip above the horizontal line through the hinge.

Solution. Multiply the coordinates of the unit point by $130$ to reach the tip.

\begin{aligned} x&=130\left(-\frac{16}{65}\right),\\ x&=-32. \end{aligned}

\begin{aligned} y&=130\left(\frac{63}{65}\right),\\ y&=126. \end{aligned}

The signed projection is $-32$ cm, placing the tip $32$ cm to the left. Its perpendicular height is $126$ cm.

Turn the height into area

The [base-height formula](/geometry/perimeter-and-area/area-of-triangles/) uses the perpendicular distance from the opposite vertex to the line containing the base. For a base of $20$ cm and a $17$ cm side hinged at its left endpoint, try turns of $30^\circ$ , $90^\circ$ , and $150^\circ$ .

At $30^\circ$ the perpendicular height is half the slanted side, or $8.5$ cm.

At $90^\circ$ the height is the full $17$ cm.

At $150^\circ$ the tip has reflected across the vertical through the hinge, so the height returns to $8.5$ cm. The equal heights at $30^\circ$ and $150^\circ$ give equal areas with the same base.

For triangle $ABC$ , let $A$ , $B$ , and $C$ also denote the angle measures. The opposite sides have lengths $a$ , $b$ , and $c$ , respectively. Place $A$ at the left end of the horizontal base $AB$ of length $c$ . Put $C$ above the line $AB$ , with $AC$ of length $b$ . A rigid motion, and a reflection if needed, puts any triangle in this position. Angle $A$ is the included angle because the two given sides meet there.

Triangle area from two sides and the angle between them#

Let $D$ be the foot of the perpendicular from $C$ to the line $AB$ , and write $h$ for the length of $CD$ . The location of $D$ changes with the angle. In each case we find the same expression for $h$ .

Acute angle at $A$ . The projection is positive, so $D$ lies on the ray from $A$ through $B$ . It lies inside the base in the picture. If the projection reaches or goes beyond $B$ , the same right triangle $ACD$ still gives the height. Its hypotenuse is $b$ , and the leg opposite $A$ is $h$ .

Acute case shown with D between A and B. The perpendicular CD supplies the height.

\begin{aligned} \sin A&=\frac{h}{b},\\ h&=b\sin A. \end{aligned}

Right angle at $A$ . The perpendicular foot is $A$ itself. Side $AC$ gives the full height. The sine at $90^\circ$ is $1$ , so $h=b\sin A$ in this case too.

Right case. Side AC itself supplies the perpendicular height.

Obtuse angle at $A$ . The projection is negative, so $D$ lies on the extension of the base to the left of $A$ . The acute angle $\angle DAC$ , marked $\alpha$ , in the right triangle $ACD$ measures $180^\circ-A$ . Its opposite leg is the height.

Obtuse case. Extend AB to the left of A to meet the perpendicular at D.

\begin{aligned} h&=b\sin(180^\circ-A),\\ h&=b\sin A. \end{aligned}

Multiply the base $c$ by the height and take half. The area $K$ is

K=\frac12bc\sin A.

The same argument works from either of the other vertices. Use the two sides that meet at the angle you choose.

\begin{aligned} K&=\frac12ac\sin B,\\ K&=\frac12ab\sin C. \end{aligned}

Check your understanding

Angle $A$ measures $137^\circ$ , and the perpendicular from $C$ to the line $AB$ meets it at $D$ . What is the measure of $\angle DAC$ ?

137^\circ

47^\circ

90^\circ

43^\circ

Use the included angle

Worked example 3 Find an exact area from the shared vertex

Problem. In triangle $ABC$ , $AB$ measures $18$ cm, $BC$ measures $13$ cm, and angle $B$ measures $45^\circ$ . Find the exact area.

The 18 cm side and the 13 cm side share vertex B.

Solution. Both given sides end at $B$ , so use $\sin B$ . The exact value of $\sin45^\circ$ is $\tfrac{\sqrt2}{2}$ .

\begin{aligned} K&=\frac12(18)(13)\sin45^\circ,\\ K&=\frac12(18)(13)\frac{\sqrt2}{2},\\ K&=\frac{117\sqrt2}{2}\text{ cm}^2. \end{aligned}

Worked example 4 Estimate the area marked by two sight lines

Problem. From one corner of a triangular wetland plot, a surveyor measures straight horizontal distances of $23.6$ m and $15.4$ m to the other two corners. The angle between these sight lines is $127^\circ$ . Find the horizontal area of the plot to the nearest tenth of a square meter.

The surveyor measures both distances and the angle from the same corner.

Solution. The distances are sides of the triangle, and the angle is measured where they meet. Enter the sine in degree mode and round only the final area.

\begin{aligned} K&=\frac12(23.6)(15.4)\sin127^\circ,\\ K&\approx 145.1\text{ m}^2. \end{aligned}

Check your understanding

In triangle $ABC$ , you know the lengths $a$ and $c$ and the measure of angle $B$ . Which expression gives the area?

\tfrac12ac\sin B

\tfrac12ac\cos B

ac\sin B

\tfrac12ac\sin A

Worked example 5 Compare two hinge settings without calculating sine

Problem. In two triangles, sides of lengths $12$ units and $19$ units meet at a hinge. One included angle is $28^\circ$ and the other is $152^\circ$ . Keep the $19$ -unit side horizontal and directed right from the hinge in each triangle. Compare their areas and the signed horizontal projections of their $12$ -unit sides.

Solution. The angles are supplementary. Their sines agree, so the areas agree because the two side lengths also agree.

\sin152^\circ=\sin28^\circ.

Their cosines have opposite signs. Multiplying by $12$ preserves that relationship.

12\cos152^\circ=-12\cos28^\circ.

The projection at $28^\circ$ points right, and the projection at $152^\circ$ points left by the same distance.

For fixed side lengths, a particular positive area fixes the height of the moving tip. On the unit semicircle, a horizontal line below the top meets the arc in two reflected points. At the top it meets the arc in one point. That top point gives the greatest area.

Check your understanding

Two fixed positive side lengths form a hinge. At which included angle is the triangle's area greatest?

45^\circ

100^\circ

90^\circ

135^\circ

Triangle Area with Sine Advanced. This lesson goes beyond core Geometry. You can skip it.

Learning goals

From a right triangle to a semicircle

What a reflection preserves

Scale the projections

Turn the height into area

Use the included angle

Common mistakes

Practice

Multiple Choice Questions (MCQ)

Core practice

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