Trigonometry Review Problems

Right Triangles

Find the indicated parts in each right triangle where the right angle is at C. Round lengths to the nearest tenth and angle measures to the nearest minute.

1. $B = 52°10', c = 61.8 \text{ cm}$; find $b$.

• Using the sine function: $\sin(B) = \frac{b}{c}$
• $b = c \cdot \sin(B) = 61.8 \cdot \sin(52°10') \approx 48.8 \text{ cm}$

1. $a = 374 \text{ ft}, b = 298 \text{ ft}$; find $B$.

• Using the tangent function: $\tan(B) = \frac{b}{a}$
• $B = \arctan(\frac{b}{a}) = \arctan(\frac{298}{374}) \approx 38.5° \approx 38°30'$

1. $A = 72.8°, c = 47.9 \text{ m}$; find $a$.

• Using the sine function: $\sin(A) = \frac{a}{c}$
• $a = c \cdot \sin(A) = 47.9 \cdot \sin(72.8°) \approx 45.7 \text{ m}$

Triangle ABC

Find the indicated parts of triangle ABC. Round answers as indicated in I.

4. $B = 72°10', b = 8370 \text{ m}, A = 21°20'$; Find $a$.

• Using the Law of Sines: $\frac{a}{\sin(A)} = \frac{b}{\sin(B)}$
• $a = \frac{b \cdot \sin(A)}{\sin(B)} = \frac{8370 \cdot \sin(21°20')}{\sin(72°10')} \approx 3200 \text{ m}$

4. $C = 107°20', a = 8.74 \text{ ft}, b = 10.3 \text{ ft}$; Find $c$.

• Using the Law of Cosines: $c^2 = a^2 + b^2 - 2ab \cos(C)$
• $c = \sqrt{a^2 + b^2 - 2ab \cos(C)} = \sqrt{8.74^2 + 10.3^2 - 2 \cdot 8.74 \cdot 10.3 \cdot \cos(107°20')} \approx 14.8 \text{ ft}$

4. $A = 42°20', a = 112 \text{ yd}, b = 138 \text{ yd}$; Find all possible values of $B$.

• Using the Law of Sines: $\frac{\sin(A)}{a} = \frac{\sin(B)}{b}$
• $\sin(B) = \frac{b \cdot \sin(A)}{a} = \frac{138 \cdot \sin(42°20')}{112} \approx 0.827$
• $B_1 = \arcsin(0.827) \approx 55.8° \approx 55°50'$
• $B<em>2 = 180° - B</em>1 \approx 180° - 55.8° = 124.2° \approx 124°10'$
• Since A + B_2 = 42°20' + 124°10' = 166°30' < 180°, both values of $B$ are possible.

4. $a = 7.02 \text{ m}, b = 3.45 \text{ m}, c = 9.17 \text{ m}$; Find $C$.

• Using the Law of Cosines: $c^2 = a^2 + b^2 - 2ab \cos(C)$
• $\cos(C) = \frac{a^2 + b^2 - c^2}{2ab} = \frac{7.02^2 + 3.45^2 - 9.17^2}{2 \cdot 7.02 \cdot 3.45} \approx -0.744$
• $C = \arccos(-0.744) \approx 138.0° \approx 138°0'$

4. Find $B$ in #7.

*Using the Law of Cosines: $b^2 = a^2 + c^2 - 2ac \cos(B)$
*$\cos(B) = \frac{a^2 + c^2 - b^2}{2ac} = \frac{7.02^2 + 9.17^2 - 3.45^2}{2 \cdot 7.02 \cdot 9.17} \approx 0.943$
*$B = \arccos(0.943) \approx 19.5° \approx 19°30'$

Area of Triangles

Find the area of each triangle to the nearest tenth.

9. $c = 804 \text{ ft}, b = 1270 \text{ ft}, A = 28°40'$

• Area $= \frac{1}{2}bc \sin(A) = \frac{1}{2} \cdot 1270 \cdot 804 \cdot \sin(28°40') \approx 244000 \text{ ft}^2$

9. $b = 3.47 \text{ m}, c = 2.99 \text{ m}, A = 142°10'$

• Area $= \frac{1}{2}bc \sin(A) = \frac{1}{2} \cdot 3.47 \cdot 2.99 \cdot \sin(142°10') \approx 3.1 \text{ m}^2$

9. $a = 0.806 \text{ ft}, b = 0.917 \text{ ft}, c = 0.732 \text{ ft}$

• Using Heron's formula: $s = \frac{a+b+c}{2} = \frac{0.806 + 0.917 + 0.732}{2} \approx 1.2275$
• Area $= \sqrt{s(s-a)(s-b)(s-c)} = \sqrt{1.2275(1.2275-0.806)(1.2275-0.917)(1.2275-0.732)} \approx 0.28 \text{ ft}^2$

Applications

Answer each. Round answers as indicated in I and III.

12. In $\triangle ABC$, $m\angle A = 10°$, $m\angle C = 150°$, and the length of $BC$ is 7 mm. Find the length of $AC$.

• Find angle B: $m\angle B = 180° - (10° + 150°) = 20°$
• Using the Law of Sines: $\frac{AC}{\sin(B)} = \frac{BC}{\sin(A)}$
• $AC = \frac{BC \cdot \sin(B)}{\sin(A)} = \frac{7 \cdot \sin(20°)}{\sin(10°)} \approx 13.8 \text{ mm}$

12. In $\triangle RST$, $m\angle S = 24°$, $s = 18 \text{ cm}$, and $t = 26 \text{ cm}$. Find the two possible measures of $\angle T$.

• Using the Law of Sines: $\frac{\sin(T)}{t} = \frac{\sin(S)}{s}$
• $\sin(T) = \frac{t \cdot \sin(S)}{s} = \frac{26 \cdot \sin(24°)}{18} \approx 0.587$
• $T_1 = \arcsin(0.587) \approx 35.9° \approx 35°54'$
• $T<em>2 = 180° - T</em>1 \approx 180° - 35.9° = 144.1°$
• Check if both angles are possible: 24 + 144.1 = 168.1 < 180, so both are possible

12. The sides of a triangle are 8.2 cm, 6.8 cm, and 10.5 cm. Find the measure of the smallest angle of the triangle.

• The smallest angle is opposite the shortest side, which is 6.8 cm. Let this angle be $B$.
• Using the Law of Cosines: $b^2 = a^2 + c^2 - 2ac \cos(B)$
• $\cos(B) = \frac{a^2 + c^2 - b^2}{2ac} = \frac{8.2^2 + 10.5^2 - 6.8^2}{2 \cdot 8.2 \cdot 10.5} \approx 0.777$
• $B = \arccos(0.777) \approx 39.1° \approx 39°06'$

12. In a triangle an angle of measure 73° is included between sides of length 7 inches and 5 inches. Find the length of the third side.

• Using the Law of Cosines: $c^2 = a^2 + b^2 - 2ab \cos(C)$
• $c = \sqrt{7^2 + 5^2 - 2 \cdot 7 \cdot 5 \cdot \cos(73°)} \approx 7.3 \text{ inches}$

12. Find the area if $\triangle ABC$ given $m\angle A = 37°$, $m\angle B = 62°$, and $a = 47 \text{ ft}$.

• Find angle C: $C = 180° - (37° + 62°) = 81°$
• Using the Law of Sines: $\frac{a}{\sin(A)} = \frac{b}{\sin(B)}$
• $b = \frac{a \cdot \sin(B)}{\sin(A)} = \frac{47 \cdot \sin(62°)}{\sin(37°)} \approx 69.4 \text{ ft}$
• Area $= \frac{1}{2}ab \sin(C) = \frac{1}{2}(47)(69.4) \sin(81°) \approx 1615 \text{ ft}^2$

12. Find the area of an isosceles triangle in which the lengths of the equal sides are 2.4 cm and one angle measures 118°.

• The $118°$ angle is between the two equal sides.
  *Area = $\frac{1}{2}ab\sin(C)$ where a and b are the equal sides.
  *Area = $\frac{1}{2}(2.4)(2.4)\sin(118°) \approx 2.5 \text{ cm}^2$

12. The hour hand of a clock is 4 inches long. The minute hand is 6 inches long. At 2:00 how far apart are the tips of the hands?

• The angle between the hands at 2:00 is $\frac{2}{12} \cdot 360° = 60°$.
• Using the Law of Cosines: $d^2 = 4^2 + 6^2 - 2 \cdot 4 \cdot 6 \cdot \cos(60°)$
• $d = \sqrt{16 + 36 - 48 \cdot 0.5} = \sqrt{28} \approx 5.3 \text{ inches}$

12. The sides of a rhombus are 5 cm each and one diagonal is 6 cm long. Find the area of the rhombus.

• The diagonals of a rhombus bisect each other at right angles. Half of one diagonal is 3 cm.
• Using Pythagorean theorem to find half of the other diagonal $\sqrt{5^2 - 3^2} = 4$
• The other diagonal is 8 cm. Area $= \frac{1}{2} d<em>1 d</em>2 = \frac{1}{2}(6)(8) = 24 \text{ cm}^2$

12. A team of surveyors must find the height of a mountain peak. Point A is known to be 6500 ft. above sea level. AB is a 600-foot stretch of level ground at the base of the mountain with A farther from the base than B. The peak can be seen from both A and B. The angle of elevation of the peak from A is a 20° angle, and from B is a 35° angle. How far above sea level is the mountain peak?

• Let h be the height from point A to the mountain peak. Let x be the horizontal distance from point A to the point directly below the peak.
• $\tan(20°) = \frac{h}{x}$ and $\tan(35°) = \frac{h}{x - 600}$
• $x = \frac{h}{\tan(20°)}$ and $x - 600 = \frac{h}{\tan(35°)}$
• $\frac{h}{\tan(20°)} - 600 = \frac{h}{\tan(35°)}$
• $h(\frac{1}{\tan(20°)} - \frac{1}{\tan(35°)}) = 600$
• $h = \frac{600}{\frac{1}{\tan(20°)} - \frac{1}{\tan(35°)}}=\frac{600}{2.747-1.428} \approx 457.7 \text{ ft}$
• The height of the mountain peak above sea level is $6500 + 457.7 \approx 6957.7 \text{ ft}$

Solving Triangles

Solve the triangle - find all missing sides and angles.

21. In $\triangle ABC$, $a = 25$, $b = 37$, and $A = 27°$.

• Using Law of Sines: $\frac{\sin A}{a} = \frac{\sin B}{b}$
• $\sin B = \frac{b \sin A}{a} = \frac{37 \sin 27°}{25} \approx 0.669$
• $B = \arcsin(0.669) \approx 41.98°$ or $B = 180° - 41.98° = 138.02°$
• If $B \approx 41.98°$, then $C = 180° - (27° + 41.98°) = 111.02°$ . $c = \frac{a \sin C}{\sin A} = \frac{25 \sin 111.02°}{\sin 27°} \approx 51.6$
• If $B \approx 138.02°$, then $C = 180° - (27° + 138.02°) = 14.98°$. $c = \frac{a \sin C}{\sin A} = \frac{25 \sin 14.98°}{\sin 27°} \approx 14.2$

21. In $\triangle ABC$, $a = 27$, $b = 28$, and $A = 109°$.

• Using the Law of Sines: $\frac{\sin(A)}{a} = \frac{\sin(B)}{b}$
• $\sin(B) = \frac{b \cdot \sin(A)}{a} = \frac{28 \cdot \sin(109°)}{27} \approx 0.977$
• $B = \arcsin(0.977) \approx 77.7°$
• $C = 180° - (109° + 77.7°) = -6.7$ which is impossible, thus no triangles exist.

21. In $\triangle ABC$, $a = 90$, $b = 70$, and $A = 80°$.

• Using the Law of Sines: $\frac{\sin(A)}{a} = \frac{\sin(B)}{b}$
• $\sin(B) = \frac{b \cdot \sin(A)}{a} = \frac{70 \cdot \sin(80°)}{90} \approx 0.766$
• $B = \arcsin(0.766) \approx 50.0°$
• $C = 180° - (80° + 50°) = 50°$
• $\frac{\sin(A)}{a} = \frac{\sin(C)}{c}$
• $c = \frac{a \cdot \sin(C)}{\sin(A)} = \frac{90 \cdot \sin(50°)}{\sin(80°)} \approx 70.0$

Number of Solutions

Find the number of solutions for the given data. Do not solve the triangle.

24. $A = 43°$, $a = 3$, $b = 4$

• $\sin B = \frac{b \sin A}{a} = \frac{4 \sin 43°}{3} \approx 0.909$. Since 0.909<1, there are two solutions.

24. $C = 95°$, $b = 5$, $c = 6$

• Since $C$ is obtuse and c > b, there is one solution.

24. $B = 91°$, $c = 12$, $b = 10$

• Since $B$ is obtuse and b < c, there are no solutions.

24. $b = 12$, $C = 4$, $c = 10$

• Can't be determined, angle C should be bigger than 4.

24. $A = 96°$, $a = 6$, $b = 6$

• Since $A$ is obtuse and $a = b$, there are no solutions.

24. $B = 46°$, $a = 50$, $b = 6$

• \sin A = \frac{a \sin B}{b} = \frac{50 \sin 46°}{6} \approx 5.99 > 1 there are no solutions.

24. $A = 93°$, $a = 10$, $b = 6$

• A is obtuse and a>b - so there is one solution

24. $a = 5$, $b = 2$, $C = 102°$

• Angle C is obtuse, side c should be the largest, we are not given side c though, but we also do not need it. Only need to focus on angle and side c in this case, and since it's obtuse, there can be one solution only. Because it is side-side-angle, that only gives one triangle. So there's one solution.