Basic Trigonometry Problems and Step-by-Step Answers

The angle formed moves clockwise, so its sign is negative, namely $-30^{\circ}.$ Since one full rotation equals $360^{\circ},$ then $-30^{\circ}$ is equivalent to $(360-30)^{\circ} = 330^{\circ}.$
Therefore, the angle measure is $\boxed{330^{\circ}}.$
(Answer D)

Recall that $\pi~\text{rad} = 180^{\circ}.$ Thus,
$\begin{aligned} \dfrac34\pi~\text{rad} & = \dfrac{3}{\cancel{4}} \times \cancelto{45}{180}^{\circ} \\ & = 3 \times 45^{\circ} = 135^{\circ}. \end{aligned}$
Therefore, the angle measure $\dfrac34\pi~\text{rad}$ is equal to $\boxed{135^{\circ}}.$
(Answer C)

Recall that $1^{\circ} = \dfrac{\pi}{180}~\text{rad}.$ Thus,
$\begin{aligned} 72^{\circ} & = \cancelto{2}{72} \times \dfrac{\pi}{\cancelto{5}{180}}~\text{rad} \\ & = \dfrac25\pi~\text{rad}. \end{aligned}$
Therefore, the angle measure $72^{\circ}$ is equal to $\boxed{\dfrac25\pi~\text{rad}}.$
(Answer B)

Based on the figure above, the trigonometric ratios for sine, cosine, and tangent of angles $\alpha$ and $\beta$ are as follows.
$\begin{aligned} \sin \alpha & = \dfrac{\text{opp}}{\text{hyp}} = \dfrac{BC}{AB} \\ \cos \alpha & = \dfrac{\text{adj}}{\text{hyp}} = \dfrac{AC}{AB} \\ \tan \alpha & = \dfrac{\text{opp}}{\text{adj}} = \dfrac{BC}{AC} \\ \sin \beta & = \dfrac{\text{opp}}{\text{hyp}} = \dfrac{AC}{AB} \\ \cos \beta & = \dfrac{\text{adj}}{\text{hyp}} = \dfrac{BC}{AB} \\ \tan \beta & = \dfrac{\text{opp}}{\text{adj}} = \dfrac{AC}{BC} \end{aligned}$
Therefore, among the five given statements (choices), the incorrect statement is option D.

Using the Pythagorean theorem, the length of $c = AB$ can be determined as follows.
$\begin{aligned} c & = \sqrt{a^2+b^2} \\ & = \sqrt{(\sqrt3)^2+1^2} \\ & = \sqrt4=2 \end{aligned}$
The cosine of an angle is the ratio between the side adjacent to the angle and the hypotenuse of a right triangle. Therefore, $\boxed{\cos \alpha = \dfrac{b}{c} = \dfrac{1}{2}}.$
(Answer D)

It is known that $x = y = -2\sqrt2.$ The polar coordinates are written in the form $(r, \theta)$ with
$\begin{aligned} r & = \sqrt{x^2+y^2} \\ & = \sqrt{(-2\sqrt2)^2+(-2\sqrt2)^2} \\ & = \sqrt{8+8} = 4. \end{aligned}$
and
$\begin{aligned} & \tan \theta = \dfrac{y} {x} = \dfrac{-2\sqrt2}{-2\sqrt2} = 1 \\ & \Rightarrow \theta = 45^{\circ} \lor 225^{\circ}. \end{aligned}$
Since point $A$ lies in quadrant III (both $x$ and $y$ are negative), then $\theta = 225^{\circ}.$ Therefore, the polar coordinates of $A(-2\sqrt2,-2\sqrt2)$ are $\boxed{(4, 225^{\circ})}.$
(Answer E)

First, sketch triangle $KLM$ on the Cartesian coordinate system as follows.

It can be seen that triangle $KLM$ is a right triangle (at $L$).
From the figure above, it is known that $KL = 3 -(-5) = 8;$ $KM = 4 -(-2) = 6.$
Using the Pythagorean theorem, we obtain
$\begin{aligned} LM & = \sqrt{KL^2 + KM^2} \\ & = \sqrt{8^2 + 6^2} \\ & = \sqrt{64+36} = \sqrt{100} = 10. \end{aligned}$
Therefore,
$\begin{aligned} \cos L & = \dfrac{KL}{LM} = \dfrac{8}{10} = \dfrac45 \\ \tan M & = \dfrac{KL}{KM} = \dfrac86 = \dfrac43. \end{aligned}$
Therefore, the values of $\cos L$ and $\tan M$ respectively are $\boxed{\dfrac45}$ and $\boxed{\dfrac43}.$
(Answer E)

First, sketch triangle $KLM$ on the Cartesian coordinate system as follows.
It can be seen that triangle $PQR$ is a right triangle (at $P$).
Without analyzing further about the side lengths of triangle $PQR$, we can actually directly calculate the value of $\dfrac{3 \sec R}{\csc Q}$ as follows by remembering that secant is the reciprocal of cosine (hyp/adj), while cosecant is the reciprocal of sine (hyp/opp). Therefore,
$\dfrac{3 \sec R}{\csc Q} = \dfrac{3 \times \cancel{\dfrac{QR}{PR}}}{\cancel{\dfrac{QR}{PR}}} = 3.$
(Answer C)

The cosine of an angle is the ratio of the side adjacent to the angle to the hypotenuse in a right triangle.
Therefore, $\cos A = \dfrac{3}{4} = \dfrac{AB}{AC}.$
Suppose $AB = 3$ and $AC = 4,$ then by using the Pythagorean theorem, we obtain
$\begin{aligned} BC & = \sqrt{AC^2 -AB^2} \\ & = \sqrt{(4)^2-(3)^2} = \sqrt{7}. \end{aligned}$
The cotangent of an angle is the ratio of the side adjacent to the angle to the side opposite the angle in a right triangle.
Therefore, $\cot A = \dfrac{AB}{BC} = \dfrac{3}{\sqrt7} = \dfrac37\sqrt7.$
Therefore, the value of $\boxed{\cot A = \dfrac37\sqrt7}.$
(Answer B)

Since $P$ is an acute angle, all trigonometric ratios are positive.
The tangent of an angle is the ratio of the side opposite the angle to the side adjacent to the angle in a right triangle.
Therefore, $\tan P = \dfrac{5\sqrt{11}}{11} = \dfrac{\text{opp}}{\text{adj}}.$
Suppose $\text{opp} = 5\sqrt{11}$ and $\text{adj} = 11,$ then using the Pythagorean theorem, the hypotenuse length is obtained as follows:
$\begin{aligned} \text{hyp} & = \sqrt{(\text{opp})^2 + (\text{adj})^2} \\ & = \sqrt{(5\sqrt{11})^2 + (11)^2} \\ &= \sqrt{275 + 121} \\ & = \sqrt{396} \\ & = \sqrt{36 \times 11} = 6\sqrt{11}. \end{aligned}$
The sine of an angle is the ratio of the side opposite the angle to the hypotenuse in a right triangle.
Therefore, $\sin P = \dfrac{\text{opp}}{\text{hyp}} = \dfrac{5\cancel{\sqrt{11}}}{6\cancel{\sqrt{11}}} = \dfrac56.$
Therefore, the value of $\boxed{\sin P = \dfrac56}.$
(Answer C)

Observe the following sketch of right triangle $ABC$.
Since $\sin B = p = \dfrac{p}{1} = \dfrac{AC}{AB}$, we may suppose that $AC = p$ and $AB = 1,$ so by using the Pythagorean theorem, we obtain
$\begin{aligned} BC & = \sqrt{AB^2 -AC^2} \\ & = \sqrt{(1)^2-p^2} \\ & = \sqrt{1-p^2}. \end{aligned}$
Thus, $\tan B = \dfrac{\text{opp}}{\text{adj}} = \dfrac{AC}{BC} = \dfrac{p}{\sqrt{1-p^2}}.$
Therefore, the value of $\boxed{\tan B = \dfrac{p}{\sqrt{1-p^2}}}.$
(Answer A)

The cosine of an angle is the ratio of the side adjacent to the angle to the hypotenuse in a right triangle.
Therefore, $\cos K = \dfrac{1}{a} = \dfrac{KL}{KM}.$
Suppose $KL = 1$ and $KM = a$, then using the Pythagorean theorem, we obtain
$\begin{aligned} LM & = \sqrt{KM^2 -KL^2} \\ & = \sqrt{a^2-(1)^2} \\ & = \sqrt{a^2-1}. \end{aligned}$
The sine of an angle is the ratio of the side opposite the angle to the hypotenuse in a right triangle, while the tangent of an angle is the ratio of the side opposite the angle to the adjacent side in a right triangle. Therefore,
$\begin{aligned} \sin K \tan K & = \dfrac{LM}{KM} \times \dfrac{LM}{KL} \\ & = \dfrac{\sqrt{a^2-1}}{a} \times \dfrac{\sqrt{a^2-1}}{1} \\ & = \dfrac{a^2-1}{a}. \end{aligned}$
Therefore, the value of $\boxed{\sin K \tan K = \dfrac{a^2-1}{a}}.$
(Answer B)

Without paying attention to the given right triangle figure, the length of the side opposite angle $\theta$ can be calculated using the Pythagorean theorem.
In this case, since $\cos \theta = \dfrac23$, let $\text{adj} = 2$ and $\text{hyp} = 3$ so that $\text{opp} = \sqrt{3^2 -2^2} = \sqrt5.$
Thus, $\sin \theta = \dfrac{\text{opp}}{\text{hyp}} = \dfrac{\sqrt5}{3}.$ Based on the given figure, it must be that $\sin \theta=\dfrac{5}{x}.$ Consequently,
$\dfrac{\sqrt5}{3} = \dfrac{5}{x} \Leftrightarrow \dfrac{\cancel{5}}{3\sqrt5} = \dfrac{\cancel{5}}{x}.$
Therefore, the value of $x$ is $\boxed{3\sqrt5}.$
(Answer A)

Since $\alpha$ is in quadrant I, all trigonometric ratios are positive.
It is known that $\tan \alpha = \dfrac{\text{opp}}{\text{adj}} = \dfrac{1}{a}$ so that we may suppose the length of the side opposite the angle is $\text{opp} = 1$ and the length of the adjacent side is $\text{adj} = a.$
Thus, the hypotenuse length of the right triangle is
$\begin{aligned} \text{hyp} & = \sqrt{(\text{opp})^2+(\text{adj})^2} \\ & = \sqrt{1^2+a^2}. \end{aligned}$
Therefore, we obtain
$$\begin{aligned} \cos \alpha -\dfrac{1}{\sin \alpha} & = \dfrac{\text{adj}}{\text{hyp}} -\dfrac{\text{hyp}}{\text{opp}} \\ & = \dfrac{a}{\sqrt{1+a^2}}- \dfrac{\sqrt{1+a^2}}{1} \\ & = \dfrac{a -(a^2+1)}{\sqrt{a^2+1}} \\ & = \dfrac{-a^2+a-1}{\sqrt{a^2+1}}. \end{aligned}$$Therefore, the value of $\boxed{\cos \alpha -\dfrac{1}{\sin \alpha} = \dfrac{-a^2+a-1}{\sqrt{a^2+1}}}.$
(Answer E)

Observe the following sketch.
The sine of an angle is the ratio of the side opposite the angle to the hypotenuse in a right triangle. Thus, it can be written as
$\begin{aligned} \sin M & = \dfrac23 \\ \dfrac{KL}{KM} & = \dfrac23 \\ \dfrac{\sqrt{20}}{KM} & = \dfrac23 \\ KM & = \dfrac{3\sqrt{20}}{2} = \dfrac{3 \times \cancel{2}\sqrt5}{\cancel{2}} \\ & = 3\sqrt5~\text{cm}. \end{aligned}$
Therefore, the side length $\boxed{KM = 3\sqrt5~\text{cm}}.$
(Answer B)

Observe the following sketch.
Since the area of triangle $DEF$ is $9~\text{cm}^2,$ by using the triangle area formula, we obtain
$\begin{aligned} L_{\triangle DEF} & = \dfrac{DE \times DF}{2} \\ 9 & = \dfrac{3 \times DF}{2} \\ DF & = \dfrac{9 \times 2}{3} = 6~\text{cm}. \end{aligned}$
Next, using the Pythagorean theorem, we obtain
$\begin{aligned} EF & = \sqrt{DE^2 + DF^2} \\ & = \sqrt{3^2+6^2} \\ & = \sqrt{9+36} = \sqrt{45} = 3\sqrt5~\text{cm}. \end{aligned}$
Thus,
$\begin{aligned} \cos E & = \dfrac{\text{adj}}{\text{hyp}} = \dfrac{DE}{EF} \\ & = \dfrac{\cancel{3}}{\cancel{3}\sqrt5} = \dfrac15\sqrt5~\text{cm}. \end{aligned}$
Therefore, the value of $\boxed{\cos E = \dfrac15\sqrt5~\text{cm}}.$
(Answer A)

Notice that by using the Pythagorean theorem, we obtain the following two equations.
$\begin{cases} c^2 & = a^2 -b^2 \\ e^2 & = f^2 + g^2 \end{cases}$
Thus, we obtain
$\sin^2 \theta = \dfrac{c^2}{e^2} = \dfrac{a^2-b^2}{f^2+g^2}.$
(Answer E)

To obtain the form of $\tan x$, note that $\dfrac{\sin x}{\cos x} = \tan x$, so we need to divide both the numerator and denominator by $\cos x.$ Thus,
$$\begin{aligned} \dfrac{5 \sin x + 6 \cos x}{2 \cos x -3 \sin x} & = \dfrac{\dfrac{5 \sin x}{\cos x} + \dfrac{6 \cos x}{\cos x}}{\dfrac{2 \cos x}{\cos x} -\dfrac{3 \sin x}{\cos x}} \\ & = \dfrac{5 \tan x + 6}{2 -3 \tan x} \\ & = \dfrac{5\left(-\dfrac23\right) + 6}{2 -3\left(-\dfrac23\right)} \\ & = \dfrac{\frac83}{4} = \dfrac{8}{12} = \dfrac23. \end{aligned}$$Thus, the value of $\boxed{\dfrac{5 \sin x + 6 \cos x}{2 \cos x -3 \sin x} = \dfrac23}.$
(Answer D)

Consider the right triangle $ABC.$
Using the cosine ratio, we have
$\begin{aligned} \cos \beta & = \dfrac{AB}{BC} = \dfrac{AB}{a} \\ AB & = a \cos \beta. \end{aligned}$
Now consider the right triangle $ABD$ (right angle at $D$).
Using the sine ratio, we have
$\begin{aligned} \sin \beta & = \dfrac{AD}{AB} \\ AD & = AB \sin \beta \\ AD & = (a \cos \beta) \sin \beta = a \sin \beta \cos \beta. \end{aligned}$
Thus, the length of the altitude is $\boxed{AD = a \sin \beta \cos \beta}.$
(Answer B)

Consider the right triangle $ABD.$
The sine value of angle alpha is given by
$\sin \alpha = \dfrac{AD}{BD} \Leftrightarrow BD = \dfrac{AD}{\sin \alpha} = \dfrac{p}{\sin \alpha}.$
Consider the right triangle $BCD.$
The cosine value of angle beta is given by $\cos \beta = \dfrac{BC}{BD}$, so we obtain
$\begin{aligned} BC & = \cos \beta \cdot BD \\ & = \cos \beta \cdot \dfrac{p}{\sin \alpha} \\ & = \dfrac{p \cos \beta}{\sin \alpha}. \end{aligned}$
Thus, the length of the side is $\boxed{BC = \dfrac{p \cos \beta}{\sin \alpha}}.$
(Answer D)

Answer a)
From the given figure, it is known that the lengths of the side adjacent to angle alpha and the hypotenuse in the right triangle are respectively $\text{sa} = 12$ and $\text{mi} = 15.$
Using the Pythagorean theorem, the length of the side opposite angle $\alpha$ is obtained, namely
$\begin{aligned} \text{de} & = \sqrt{15^2-12^2} \\ & = \sqrt{225-144} = \sqrt{81} = 9. \end{aligned}$
Therefore, we obtain
$\begin{aligned} \sin \alpha & = \dfrac{\text{de}}{\text{mi}} = \dfrac{9}{15} = \dfrac35 \\ \cos \alpha & = \dfrac{\text{sa}}{\text{mi}} = \dfrac{12}{15} = \dfrac45 \\ \tan \alpha & = \dfrac{\text{de}}{\text{sa}} = \dfrac{9}{12} = \dfrac34 \\ \csc \alpha & = \dfrac{\text{mi}}{\text{de}} = \dfrac{15}{9} = \dfrac53 \\ \sec \alpha & = \dfrac{\text{mi}}{\text{sa}} = \dfrac{15}{12} = \dfrac54 \\ \cot \alpha & = \dfrac{\text{sa}}{\text{de}} = \dfrac{12}{9} = \dfrac43. \end{aligned}$
Answer b)
From the given figure, it is known that the lengths of the opposite side and adjacent side to angle alpha in the right triangle are respectively $\text{de} = 12$ and $\text{sa} =5.$ Using the Pythagorean theorem, the length of the hypotenuse is obtained, namely
$\begin{aligned} \text{mi} & = \sqrt{12^2+5^2} \\ & = \sqrt{144+25} = \sqrt{169} = 13. \end{aligned}$
Therefore, we obtain
$\begin{aligned} \sin \alpha & = \dfrac{\text{de}}{\text{mi}} = \dfrac{12}{13} \\ \cos \alpha & = \dfrac{\text{sa}}{\text{mi}} = \dfrac{5}{13} \\ \tan \alpha & = \dfrac{\text{de}}{\text{sa}} = \dfrac{12}{5} \\ \csc \alpha & = \dfrac{\text{mi}}{\text{de}} = \dfrac{13}{12} \\ \sec \alpha & = \dfrac{\text{mi}}{\text{sa}} = \dfrac{13}{5} \\ \cot \alpha & = \dfrac{\text{sa}}{\text{de}} = \dfrac{5}{12}.\end{aligned}$
Answer c)
Redraw the figure by labeling its vertices as shown below.
The length of $AC$ can be determined using the Pythagorean theorem, namely
$\begin{aligned} AC & = \sqrt{AD^2 + CD^2} \\ & = \sqrt{24^2+7^2} \\ & = \sqrt{576+49} = \sqrt{625} = 25. \end{aligned}$
The length of $BC$ can also be determined using the Pythagorean theorem, namely
$\begin{aligned} BC & = \sqrt{AC^2 -AB^2} \\ & = \sqrt{25^2-20^2} \\ & = \sqrt{625-400} = \sqrt{225} = 15. \end{aligned}$
From here, it is known that the lengths of the side opposite angle alpha, the side adjacent to angle alpha, and the hypotenuse in $\triangle ABC$ are respectively
$\text{de} = 15~~~\text{sa} = 20~~~\text{mi} = 25.$
Therefore, we obtain
$\begin{aligned} \sin \alpha & = \dfrac{\text{de}}{\text{mi}} = \dfrac{15}{25} = \dfrac35 \\ \cos \alpha & = \dfrac{\text{sa}}{\text{mi}} = \dfrac{20}{25} = \dfrac45 \\ \tan \alpha & = \dfrac{\text{de}}{\text{sa}} = \dfrac{15}{20} = \dfrac34 \\ \csc \alpha & = \dfrac{\text{mi}}{\text{de}} = \dfrac{25}{15} = \dfrac53 \\ \sec \alpha & = \dfrac{\text{mi}}{\text{sa}} = \dfrac{25}{20} = \dfrac54 \\ \cot \alpha & = \dfrac{\text{sa}}{\text{de}} = \dfrac{20}{15} = \dfrac43. \end{aligned}$

Observe the following sketch.
Since $\sin L = 0,\!28,$ we get
$\begin{aligned} \sin L & = \dfrac{28}{100} \\ \dfrac{KM}{ML} & = \dfrac{7}{25}. \end{aligned}$
Suppose $KM = 7$ and $ML = 25,$ then using the Pythagorean theorem, we obtain
$\begin{aligned} KL & = \sqrt{ML^2-KM^2} \\ & = \sqrt{25^2-7^2} \\ & = \sqrt{625-49} = \sqrt{576} = 24. \end{aligned}$
Answer a)
$\tan L = \dfrac{KM}{KL} = \dfrac{7}{24}.$
Answer b)
$\tan M = \dfrac{KL}{KM} = \dfrac{24}{7}.$

Based on the Pythagorean theorem, $b^2=a^2+c^2.$
Answer a)
Proof of the left-hand side.
$$\begin{aligned} \sin^2 C + \cos^2 C & = \left(\dfrac{AB}{AC}\right)^2 + \left(\dfrac{BC}{AC}\right)^2 \\ & = \left(\dfrac{c}{b}\right)^2 + \left(\dfrac{a}{b}\right)^2 \\ & = \dfrac{c^2 + a^2}{b^2} \\ &= \dfrac{b^2}{b^2} = 1 \end{aligned}$$Answer b)
Proof of the left-hand side.
$$\begin{aligned} \csc^2 C -\cot^2 C & = \left(\dfrac{AC}{AB}\right)^2- \left(\dfrac{BC}{AB}\right)^2 \\ & = \left(\dfrac{b}{c}\right)^2 -\left(\dfrac{a}{c}\right)^2 \\ & = \dfrac{b^2 -a^2}{c^2} \\ &= \dfrac{c^2}{c^2} = 1 \end{aligned}$$

The tangent of an angle is the ratio of the side opposite the angle to the side adjacent to the angle in a right triangle. In triangle $ABC$, we obtain
$$\begin{aligned} \tan \alpha & = \dfrac{BC}{AB} = \dfrac{BC}{AD + BD} = \dfrac{BC}{1 + BD} \\ BC & = \tan \alpha(1 + BD) \\ BC & = \color{red}{\tan \alpha + BD \tan \alpha} \end{aligned}$$In triangle $DBC$, we obtain
$$\begin{aligned} \tan \beta & = \dfrac{BC}{BD} \\ \tan \beta & = \dfrac{ \color{red}{\tan \alpha + BD \tan \alpha}}{BD} \\ BD \tan \beta & = \tan \alpha + BD \tan \alpha \\ BD(\tan \beta -\tan \alpha) & = \tan \alpha \\ BD & = \dfrac{\tan \alpha}{\tan \beta -\tan \alpha}~\text{cm}. \end{aligned}$$(Proven)

Observe the following sketch.
Based on the properties of a square, its two diagonals must intersect perpendicularly at the midpoint, namely point $O.$
The diagonal length $AC = BD$ can be determined using the Pythagorean theorem, for example by considering right triangle $ABD.$
$\begin{aligned} AC = BD & = \sqrt{AB^2+AD^2} \\ & = \sqrt{(6a)^2 + (6a)^2} \\ & = \sqrt{(6a)^2(1+1)} \\ & = 6a\sqrt2~\text{units} \end{aligned}$
Point $P$ lies on $AC$ with ratio $OP : PC = 1 : 2.$
Since $AC = 6a\sqrt2$, then the length $OC = 3a\sqrt2$ (half of $AC$).
Based on the ratio, we obtain
$\begin{aligned} OP & = \dfrac{1}{1+2} \times OC \\ & = \dfrac13 \times 3a\sqrt2 = a\sqrt2. \end{aligned}$
Now consider right triangle $BOP$ as illustrated below.
The length $BP$ can be determined using the Pythagorean theorem as follows.
$\begin{aligned} BP & = \sqrt{BO^2 + OP^2} \\ & = \sqrt{(3a\sqrt2)^2 + (a\sqrt2)^2} \\ & = \sqrt{18a^2 + 2a^2} = \sqrt{20a^2} \\ & = 2a\sqrt5~\text{units} \end{aligned}$
Thus,
$\begin{aligned} \sin \angle PBO & = \dfrac{OP}{BP} = \dfrac{a\sqrt2}{2a\sqrt5} \\ & = \dfrac{\sqrt2}{2\sqrt5} = \dfrac{1}{10}\sqrt{10}. \end{aligned}$
Therefore, the value of $\boxed{\sin \angle PBO = \dfrac{1}{10}\sqrt{10}}$

Observe the following trapezoid sketch.
In right triangle $AGO$, it holds that $$\cos \alpha = \dfrac{a}{1} = a.$$In right triangle $OED$, it holds that $$\cos \alpha = \dfrac{1-c}{1} \Leftrightarrow c = 1-\cos \alpha.$$In right triangle $AHD$, it holds that $$\sin \alpha = \dfrac{2b}{2} = b.$$Thus, the area of the trapezoid is
$$\begin{aligned} L & = \dfrac{AB + DC}{2} \times BC \\ & = \dfrac{(a+1) + c}{\cancel{2}} \times \cancel{2}b \\ & = \left(\cos \alpha + 1 + (1-\cos \alpha)\right) \times \sin \alpha \\ & = 2 \sin \alpha. \end{aligned}$$Therefore, the area of the trapezoid is $\boxed{2 \sin \alpha}.$

Observe the following sketch.
From the figure above, it holds that
$$\begin{aligned} \tan \alpha & = \dfrac{t}{a+b} \Leftrightarrow \color{blue}{a+b = \dfrac{t}{\tan \alpha}} && (\cdots 1) \\ \tan \beta & = \dfrac{h}{a} && (\cdots 2) \\ \tan \beta & = \dfrac{h+t}{b}. && (\cdots 3) \end{aligned}$$From Equations $(2)$ and $(3),$ we obtain
$$\begin{aligned} \tan \beta & = \dfrac{h+(h+t)}{a+b} \\ \tan \beta & = \dfrac{2h+t}{a+b} \\ \color{blue}{(a+b)} \tan \beta & = 2h+t \\ \dfrac{t}{\tan \alpha} \cdot \tan \beta & = 2h+t \\ t \left(\dfrac{\tan \beta}{\tan \alpha}-1\right) & = 2h \\ t\left(\dfrac{\tan \beta-\tan \alpha}{\tan \alpha}\right) & = 2h \\ t & = \dfrac{2h \tan \alpha}{\tan \beta-\tan \alpha}. \end{aligned}$$Since the bird’s height above the water/ground surface is equal to $t + h$, we obtain
$$\begin{aligned} \text{Bird Height} & = \dfrac{2h \tan \alpha}{\tan \beta-\tan \alpha} + h \\ & = \dfrac{2h \tan \alpha}{\tan \beta-\tan \alpha} + \dfrac{h(\tan \beta- \tan \alpha)}{\tan \beta-\tan \alpha} \\ & = \dfrac{h(2 \tan \alpha+\tan \beta-\tan \alpha)}{\tan \beta-\tan \alpha} \\ & = \dfrac{\tan \alpha+\tan \beta}{\tan \beta-\tan \alpha} \cdot h. \end{aligned}$$Therefore, the bird’s height is $\boxed{\dfrac{\tan \alpha+\tan \beta}{\tan \beta-\tan \alpha} \cdot h}.$

In the triangle, it holds that
$$\begin{aligned} \sin 37^\circ & = \dfrac{BC}{AC} = \dfrac35 \\ \cos 37^\circ & = \dfrac{AB}{AC} = \dfrac45. \end{aligned}$$Suppose the side length of the square = $s.$
Position points $D, E, F$ as shown in the figure below. Also suppose $CE = x$ and $EB = y$ so that $x + y = BC = 3.$In triangle $CDE,$ we obtain
$$\begin{aligned} \cos 37^\circ & = \dfrac{DE}{CE} \\ \dfrac45 & = \dfrac{s}{x} \\ x & = \dfrac54s. \end{aligned}$$In triangle $FBE,$ we obtain
$$\begin{aligned} \sin 37^\circ & = \dfrac{EB}{FE} \\ \dfrac35 & = \dfrac{y}{s} \\ y & = \dfrac35s. \end{aligned}$$Thus,
$$\begin{aligned} x + y & = 3 \\ \dfrac54s + \dfrac35s & = 3 \\ \dfrac{25+12}{20}s & = 3 \\ \dfrac{37}{20}s & = 3 \\ s & = \dfrac{60}{37}. \end{aligned}$$Therefore, the side length of the square is $\boxed{\dfrac{60}{37}}$