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07 Applications of Trigonometry

A practical guide to solving right and non-right triangle problems, choosing trigonometric methods, finding triangle areas, and checking mathematical models.

Build the Triangle Model

Trigonometry connects angle measures with lengths, making it possible to determine inaccessible heights, distances, directions, and areas. Use the standard triangle notation in which side aa is opposite angle AA, side bb is opposite angle BB, and side cc is opposite angle CC. The angle sum is

A+B+C=180∘.A+B+C=180^\circ.

A reliable solution begins by drawing and labeling a diagram. Identify the known quantities, the desired quantity, and whether the triangle is right or non-right. Then select a method, solve symbolically when possible, substitute numerical values, and round only at the end.

As a calculator check, use degree mode whenever the given angles are measured in degrees.

Takeaway: The diagram and the type of given information determine which formula is appropriate.

Solve Right-Triangle Problems

A contains one 90∘90^\circ angle. The side opposite that angle is the hypotenuse. Relative to an acute angle θ\theta, identify the opposite and adjacent sides before choosing a ratio.

The key are

sin⁡θ=oppositehypotenuse,cos⁡θ=adjacenthypotenuse,tan⁡θ=oppositeadjacent.\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}},\qquad \cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}},\qquad \tan\theta=\frac{\text{opposite}}{\text{adjacent}}.

The Pythagorean theorem provides another relationship:

a2+b2=c2,a^2+b^2=c^2,

where cc is the hypotenuse.

To solve a right-triangle problem:

  1. Identify the known acute angle.

  2. Label each relevant side relative to that angle.

  3. Choose the ratio containing the known and unknown sides.

  4. Substitute and solve.

  5. Check that the result has sensible units and magnitude.

For example, if a point is 5050 feet from a building and the angle of elevation to the top is 68∘68^\circ, the height hh above eye level satisfies

tan⁡68∘=h50.\tan68^\circ=\frac{h}{50}.

Thus,

h=50tan⁡68∘≈123.8 feet.h=50\tan68^\circ\approx123.8\text{ feet}.

If the observer’s eyes are 5.55.5 feet above the ground, the total building height is approximately 129.3129.3 feet.

Takeaway: Match the ratio to the side positions, not merely to the numerical values in the problem.

Use Elevation and Depression Angles

An angle of elevation is measured upward from a horizontal line to an object. An angle of depression is measured downward from a horizontal line. Because horizontal lines are parallel, an angle of depression equals the corresponding angle of elevation from the lower object, allowing a to model the situation.

Suppose the top of a 120120-foot tower has an angle of depression of 25∘25^\circ to a car. If dd is the horizontal distance from the tower’s base to the car, then

tan⁡25∘=120d.\tan25^\circ=\frac{120}{d}.

Solving gives

d=120tan⁡25∘≈257.4 feet.d=\frac{120}{\tan25^\circ}\approx257.4\text{ feet}.

The height and horizontal distance form the opposite and adjacent sides of the modeled .

Takeaway: Convert an elevation or depression situation into a labeled before selecting a trigonometric ratio.

Apply the

For a non-, first determine which side-angle information is available. The is

asin⁡A=bsin⁡B=csin⁡C.\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}.

It is useful for ASA, AAS, and SSA configurations. In an AAS example, suppose

A=42∘,B=68∘,a=10.A=42^\circ,\qquad B=68^\circ,\qquad a=10.

Find the remaining angle:

C=180∘−42∘−68∘=70∘.C=180^\circ-42^\circ-68^\circ=70^\circ.

Then use matching side-angle pairs:

b=10sin⁡68∘sin⁡42∘≈13.9,c=10sin⁡70∘sin⁡42∘≈14.0.b=\frac{10\sin68^\circ}{\sin42^\circ}\approx13.9, \qquad c=\frac{10\sin70^\circ}{\sin42^\circ}\approx14.0.

The occurs with some SSA data. If an acute angle AA, its opposite side aa, and another side bb are known, compare aa with

h=bsin⁡A.h=b\sin A.

For an acute AA:

  • If a<ha<h, no triangle exists.

  • If a=ha=h, exactly one exists.

  • If h<a<bh<a<b, two triangles may exist.

  • If a≥ba\ge b, at most one triangle exists.

When a second triangle is possible, a supplementary angle can be tested using

B2=180∘−B1,B_2=180^\circ-B_1,

but the resulting angle sum must remain less than 180∘180^\circ.

Takeaway: Match every side with its opposite angle, and always test the second SSA possibility when the measurements allow it.

Apply the

The is appropriate for SAS and SSS information. Its three forms are

a2=b2+c2−2bccos⁡A,a^2=b^2+c^2-2bc\cos A,
b2=a2+c2−2accos⁡B,b^2=a^2+c^2-2ac\cos B,
c2=a2+b2−2abcos⁡C.c^2=a^2+b^2-2ab\cos C.

For two sides of lengths 77 and 1010 enclosing an angle of 60∘60^\circ, the third side satisfies

c2=72+102−2(7)(10)cos⁡60∘=79.c^2=7^2+10^2-2(7)(10)\cos60^\circ=79.

Therefore,

c=79≈8.9.c=\sqrt{79}\approx8.9.

To find an angle from three sides, rearrange the relevant form. For example,

cos⁡C=a2+b2−c22ab.\cos C=\frac{a^2+b^2-c^2}{2ab}.

If a=8a=8, b=11b=11, and c=13c=13, then

C=cos⁡−1(82+112−1322(8)(11))≈84.8∘.C=\cos^{-1}\left(\frac{8^2+11^2-13^2}{2(8)(11)}\right)\approx84.8^\circ.

The generalizes the Pythagorean theorem. When C=90∘C=90^\circ, cos⁡90∘=0\cos90^\circ=0, so

c2=a2+b2.c^2=a^2+b^2.

Takeaway: Use the included angle in SAS problems and the side opposite the target angle when solving an SSS problem.

Find Triangle Areas

Triangle area can be found in several ways. With a base and its perpendicular height,

K=12bh.K=\frac12 bh.

With two sides and their included angle, use the :

K=12absin⁡C.K=\frac12 ab\sin C.

For sides 88 and 1111 enclosing an angle of 37∘37^\circ,

K=12(8)(11)sin⁡37∘≈26.5K=\frac12(8)(11)\sin37^\circ\approx26.5

square units.

When all three sides are known, use . First calculate the semiperimeter:

s=a+b+c2.s=\frac{a+b+c}{2}.

Then calculate

K=s(s−a)(s−b)(s−c).K=\sqrt{s(s-a)(s-b)(s-c)}.

For side lengths 1313, 1414, and 1515,

s=13+14+152=21,s=\frac{13+14+15}{2}=21,

so

K=21(21−13)(21−14)(21−15)=84K=\sqrt{21(21-13)(21-14)(21-15)}=84

square units.

Takeaway: Use a perpendicular height when available, the included-angle formula for two sides plus an angle, and for three sides.

Choose a Method and Check the Result

Use the following decision process when selecting a method:

  • One right angle with a side-angle relationship: use sin⁡\sin, cos⁡\cos, or tan⁡\tan.

  • Two angles and one side: use the .

  • Two sides and an opposite angle: use the and check for the .

  • Two sides and their included angle: use the .

  • Three sides: use the for angles or for area.

  • Two sides and their included angle when area is required: use K=12absin⁡CK=\frac12 ab\sin C.

Before finalizing a result, check that:

  • every side-angle pair is matched with its opposite;

  • a right-triangle ratio has not been applied to a non-;

  • the angle sum is 180∘180^\circ;

  • side lengths and areas are positive;

  • linear measurements use units such as feet, while areas use square units;

  • the calculator is in degree mode for degree measures; and

  • intermediate values were not rounded too early.

Final takeaway: Draw, label, select, solve, and check. This sequence makes trigonometric modeling more reliable and helps reveal incorrect formulas or unreasonable answers.