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Triangle Area with Sine: Core practice

10 practice problems for this lesson. Work on paper, use hints when you need them, and check the answer or the full solution when you are ready.

Difficulty: Advanced (beyond the core course) Advanced. This problem set goes beyond core Geometry. You can skip it.

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Problem 1 of 10

Problem 1 Read a point left of the axis

The ray for angle $\theta$ meets the upper unit semicircle at $\left(-\tfrac{15}{113},\tfrac{112}{113}\right)$ .

Find $\cos\theta$ and $\sin\theta$ . Classify $\theta$ as acute, right, or obtuse and explain how the coordinates tell you.
1. Hint 1
  Recall which coordinate each function reads.
2. Hint 2
  Compare the point with the vertical axis through the top of the semicircle.
Answer

Cosine is $-\tfrac{15}{113}$ . Sine is $\tfrac{112}{113}$ . The angle is obtuse.

Full solution

Cosine is the signed horizontal projection.
$\cos\theta=-\frac{15}{113}$

Sine is the vertical coordinate.
$\sin\theta=\frac{112}{113}$

The point lies left of the vertical axis, so its ray has turned beyond a right angle.
The angle is obtuse.

Answer

Cosine is $-\tfrac{15}{113}$ . Sine is $\tfrac{112}{113}$ . The angle is obtuse.

Key idea

The horizontal coordinate distinguishes the acute and obtuse parts of the semicircle.
Problem 2 Reflect a unit point

An acute angle $T$ has cosine $\tfrac{11}{61}$ and sine $\tfrac{60}{61}$ .

Find the coordinates of the unit point for $180^\circ-T$ . Describe the reflection that takes one unit point to the other.
1. Hint 1
  A supplementary angle completes a straight angle with the original turn.
2. Hint 2
  Reflect the radius over the vertical line through the origin and track both coordinates.
Answer

The point is $\left(-\tfrac{11}{61},\tfrac{60}{61}\right)$ . The reflection is in the vertical axis.

Full solution

The original point is $\left(\tfrac{11}{61},\tfrac{60}{61}\right)$ .
Reflect its radius in the vertical axis.

The reflected ray makes the original angle with the negative horizontal axis, so its turn from the positive axis is $180^\circ-T$ .

Reflection keeps the vertical coordinate and reverses the horizontal one.
The reflected point is $\left(-\tfrac{11}{61},\tfrac{60}{61}\right)$ .

Answer

The point is $\left(-\tfrac{11}{61},\tfrac{60}{61}\right)$ . The reflection is in the vertical axis.

Key idea

Reflection in the vertical axis preserves height and reverses the signed projection.
Problem 3 Exact area at an obtuse corner

In triangle $ABC$ , $AB$ is $27$ cm long and $AC$ is $16$ cm long. Angle $A$ measures $120^\circ$ .

Find the exact area.

Both given lengths meet at A.

Text description of this figure

The 27-centimeter horizontal side and the 16-centimeter slanted side meet at angle A, which measures 120 degrees.
1. Hint 1
  Trace the two given sides to their shared vertex.
2. Hint 2
  Use the acute supplement to get the exact sine.
Answer

$108\sqrt3\text{ cm}^2$ .

Full solution

The given sides meet at $A$ , so use the sine of $120^\circ$ .

The supplement of $120^\circ$ is $60^\circ$ .
The supplementary angles have the same sine.
$\sin120^\circ=\frac{\sqrt3}{2}$

Substitute in the area formula.
$\begin{aligned}K&=\tfrac12(27)(16)\tfrac{\sqrt3}{2},\\ K&=108\sqrt3\text{ cm}^2\end{aligned}$

Answer

$108\sqrt3\text{ cm}^2$ .

Key idea

Use the sine of the included angle, or the equal sine of its supplement.
Problem 4 Recover the length of a pennant edge

A triangular pennant has area $63\text{ ft}^2$ . Its base is $14$ ft long. At the left end, the base is extended left past the pennant. The angle between that extension and the pennant's slanted edge at this corner is $62^\circ$ .

Find the length of the slanted edge at this corner to the nearest tenth of a foot.
1. Hint 1
  The angle measured from the base extension is supplementary to the included angle.
2. Hint 2
  Write the area as one half times the base, the unknown side, and the appropriate sine.
Answer

About $10.2$ ft.

Full solution

The included angle and the $62^\circ$ exterior angle add to a straight angle.
Their sines are equal.

Let $u$ be the slanted length in feet.
Substitute the measured area and base.
$63=\tfrac12(14)u\sin62^\circ$

Divide by the coefficient of the unknown length.
Use degree mode and round the final result.
$\begin{aligned}u&=\frac9{\sin62^\circ},\\ u&\approx10.2\end{aligned}$

Answer

About $10.2$ ft.

Key idea

An exterior angle along the base extension supplies the same sine as the included angle.
Problem 5 An acute angle with an exterior foot

In triangle $ABC$ , $AB$ is horizontal and points right from $A$ . Its length is $34$ units. The side $AC$ has length $41$ units and points above the base line. Angle $A$ is acute and has cosine $\tfrac{40}{41}$ .

Locate the foot $D$ of the perpendicular from $C$ to line $AB$ relative to segment $AB$ . Find the exact area.
1. Hint 1
  Use cosine to find the horizontal coordinate of the foot measured from A.
2. Hint 2
  Use the right triangle with hypotenuse AC to find the height. Then use base AB.
Answer

The foot is $6$ units beyond $B$ . The area is $153$ square units.

Full solution

The horizontal projection of $AC$ gives the distance from $A$ to $D$ .
$\begin{aligned}AD&=41\left(\frac{40}{41}\right),\\ AD&=40\end{aligned}$

Point $B$ is $34$ units right of $A$ .
Therefore $D$ lies $6$ units beyond $B$ .

Use the right triangle $ACD$ to find the positive height $h$ .
$\begin{aligned}h^2&=41^2-40^2,\\ h&=9\end{aligned}$

The base of triangle $ABC$ still measures $34$ .
$\begin{aligned}K&=\tfrac12(34)(9),\\ K&=153\end{aligned}$

The projection exceeds the length of AB, so the altitude lands beyond B.

Answer

The foot is $6$ units beyond $B$ . The area is $153$ square units.

Key idea

An acute angle at the left endpoint places the foot to its right, possibly past the other endpoint.
Problem 6 Survey a garden in two pieces

A surveyor stands at $P$ , between endpoints $A$ and $B$ of the straight boundary of a triangular garden $ABC$ . Distances $PA$ , $PB$ , and $PC$ are $8$ m, $17$ m, and $12$ m, respectively. Angle $BPC$ measures $41^\circ$ .

Write expressions for the areas of $APC$ and $BPC$ using the sine of the same angle. Find the total area to the nearest tenth of a square meter.

Segment PC divides the garden at the point where the base measurements meet.

Text description of this figure

From P, segment PA runs 8 meters left, PB runs 17 meters right, and PC runs 12 meters up and right at 41 degrees from PB.
1. Hint 1
  Rays $PA$ and $PB$ point in opposite directions. Relate the two angles at $P$ to a straight angle.
2. Hint 2
  The two pieces use different bases but the same 12-meter side. Add their areas before rounding.
Answer

$\begin{gathered}K_{APC}=48\sin41^\circ\text{ m}^2\\ K_{BPC}=102\sin41^\circ\text{ m}^2\\ K_{ABC}\approx98.4\text{ m}^2\end{gathered}$

Full solution

Angles $APC$ and $BPC$ are supplementary.
Use $\sin41^\circ$ for both.

The first piece has sides of lengths 8 and 12 at its included angle.
$K_{APC}=48\sin41^\circ$

The second piece has sides of lengths 17 and 12 at its included angle.
$K_{BPC}=102\sin41^\circ$

Add the two areas and round the total.
$\begin{aligned}K_{ABC}&=150\sin41^\circ,\\ K_{ABC}&\approx98.4\text{ m}^2\end{aligned}$

Answer

$\begin{gathered}K_{APC}=48\sin41^\circ\text{ m}^2\\ K_{BPC}=102\sin41^\circ\text{ m}^2\\ K_{ABC}\approx98.4\text{ m}^2\end{gathered}$

Key idea

A straight baseline can give supplementary included angles in two adjoining triangles.
Problem 7 Choose the shortest support

A $32$ cm rod is fixed horizontally. A second rod hinges at its left end and can turn through any angle strictly between $0^\circ$ and $180^\circ$ . The two free ends and the hinge must enclose a triangle of area $250\text{ cm}^2$ .

Find the shortest possible length of the second rod. State the hinge angle that achieves it and justify why a shorter rod cannot work.
1. Hint 1
  First determine the perpendicular height required by the fixed base and target area.
2. Hint 2
  Compare that height with the largest possible height for a rod of a given length.
Answer

The minimum length is $\tfrac{125}{8}$ cm, achieved at $90^\circ$ .

Full solution

Solve the base-height formula for the required height in centimeters.
$\begin{aligned}250&=\tfrac12(32)h,\\ h&=\tfrac{125}{8}\end{aligned}$

The tip can rise at most the full length of the second rod.
Every shorter rod gives a height below the required value.

A rod of length $\tfrac{125}{8}$ cm standing vertically has exactly the required height.
It meets the fixed rod at $90^\circ$ .

Answer

The minimum length is $\tfrac{125}{8}$ cm, achieved at $90^\circ$ .

Key idea

The shortest hinged side for a prescribed height reaches straight up.
Problem 8 Recover two possible projections

Side $AB$ is horizontal to the right from $A$ and measures $42$ units. Vertex $C$ lies above that line at a distance of $37$ units from $A$ . Triangle $ABC$ has area $735$ square units.

Find every possible signed horizontal projection of $AC$ from $A$ . Explain why each is possible.
1. Hint 1
  The given base and area determine the vertical coordinate of C.
2. Hint 2
  Use the Pythagorean theorem with the fixed length AC. Consider both sides of the vertical line through A.
Answer

The projections are $+12$ units and $-12$ units.

Full solution

The area gives the height of $C$ above the base line.
$\begin{aligned}735&=\tfrac12(42)h,\\ h&=35\end{aligned}$

Let $x$ be the signed horizontal projection.
A right triangle with vertical leg $35$ and hypotenuse $37$ gives
$\begin{aligned}x^2&=37^2-35^2,\\ x^2&=144,\\ x&=\pm12\end{aligned}$

The two positions are reflections across the vertical line through $A$ .
Both are $37$ units from $A$ and $35$ units above the base line.
Both therefore give the specified area.

Answer

The projections are $+12$ units and $-12$ units.

Key idea

A height can determine the magnitude of a projection while leaving its sign open.
Problem 9 Change two sides while preserving area

A triangle has sides of lengths $s$ and $t$ meeting at an angle $\theta$ strictly between $0^\circ$ and $180^\circ$ . A redesign replaces the $s$ -side with one of length $\tfrac54s$ and opens the hinge to the supplement of $\theta$ .

By what factor must the $t$ -side change to preserve the area? Justify your result for every allowed $\theta$ .
1. Hint 1
  Compare the original and new sines using the supplementary-angle relationship.
2. Hint 2
  Write the new side length as a factor times the old one. Match the two products.
Answer

Multiply the $t$ -side length by $\tfrac45$ .

Full solution

The two included angles have equal positive sines.
The areas can therefore be equal exactly when the products of the given sides agree.

Let $f$ be the factor multiplying $t$ .
Write the product of the two new lengths and set it equal to the old product.
$\left(\tfrac54s\right)(ft)=st$

The original side lengths are positive, so divide by $st$ .
$\begin{aligned}\tfrac54 f&=1,\\ f&=\tfrac45\end{aligned}$

This factor restores the original product for every allowed $\theta$ , so it also restores the original area.

Answer

Multiply the $t$ -side length by $\tfrac45$ .

Key idea

For supplementary included angles, equal areas require equal products of the adjacent side lengths.
Problem 10 Compare the sides opposite two hinge settings

A hinge at $A$ keeps $AB$ horizontal to the right with length $33$ units. A $29$ -unit arm reaches $C_1$ when its angle is acute. The unit point on that ray is $\left(\tfrac{20}{29},\tfrac{21}{29}\right)$ . The arm then opens to the supplementary angle, reaching $C_2$ above the base line.

Give the two signed horizontal projections of the arm. Compare the areas of triangles $ABC_1$ and $ABC_2$ , and decide which third side, $BC_1$ or $BC_2$ , is longer. Explain the comparison using projections and heights.
1. Hint 1
  Reflect the acute unit point in the vertical axis, then scale both points to the arm length.
2. Hint 2
  Compare the horizontal distances from B to the two feet. Both right triangles for the third sides have the same height.
Answer

The projections are $+20$ and $-20$ units. The areas are equal, and $BC_2$ is longer.

Full solution

Reflection gives the second unit point $\left(-\tfrac{20}{29},\tfrac{21}{29}\right)$ .
Multiplying each unit point by the arm length gives $C_1=(20,21)$ .
$C_2=(-20,21)$

The signed projections are $+20$ and $-20$ units.
Both triangles have the same base and the same $21$ -unit height, so their areas agree.

Point $B$ is $33$ units to the right of $A$ .
The first foot lies between $A$ and $B$ , so its horizontal distance from $B$ is $13$ units.

The second foot lies left of $A$ , so its horizontal distance from $B$ is $53$ units.

Each third side is the hypotenuse of a right triangle with a $21$ -unit vertical leg.
The second right triangle has the longer horizontal leg, so the Pythagorean theorem gives $BC_2>BC_1$ .

Both heights agree. The exterior foot lies farther from B than the interior foot.

Answer

The projections are $+20$ and $-20$ units. The areas are equal, and $BC_2$ is longer.

Key idea

Equal heights preserve area with a fixed base, while opposite projections can change the third side.

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