Trigonometry Problems with Detailed Solutions

This page presents real-life trigonometry problems solved using trigonometric ratios, rotation formulas, and angles of elevation and depression.

Problem 1

A person stands 100 meters from the base of a tree. The angle of elevation to the top of the tree is \(18^\circ\). Estimate the height \(h\) of the tree.

Angle of elevation to top of a tree

Solution

Using the tangent ratio:

\[ \tan(18^\circ) = \frac{h}{100} \] \[ h = 100 \tan(18^\circ) = 32.5 \text{ m} \]

Problem 2

The angle of elevation of a hot-air balloon changes from \(25^\circ\) to \(60^\circ\) in 2 minutes. The observer is 300 meters from the takeoff point. Find the constant upward speed of the balloon.

Hot air balloon angle of elevation

Solution

\[ \tan(25^\circ) = \frac{h_1}{300} \quad (1) \] \[ \tan(60^\circ) = \frac{h_1 + h_2}{300} \quad (1) \] Equation (1) gives : \[ h_1 = 300 \tan(25^\circ) \quad (3) \] and equation (2) gives \[ h_1 + h_2 = 300 \tan(60^\circ) \quad (4) \] Subtract equation (3) from equation (4) to obtain \[ h_2 = 300\,[\tan(60^\circ) - \tan(25^\circ)] \]

Time \(= 2\) minutes \(= 120\) seconds.

\[ S = \frac{h_2}{120} = \frac{300\,[\tan(60^\circ) - \tan(25^\circ)]}{120} = 3.16 \text{ m/s} \]

Problem 3

A point \(P(x,y)\) is rotated by an angle \(a\) about the origin. Find the coordinates \((x',y')\) of the new point.

Rotation of a point about the origin

Solution

\[ x = r\cos b,\quad y = r\sin b \] \[ x' = r\cos(a+b),\quad y' = r\sin(a+b) \] Using angle addition formulas: \[ x' = x\cos a - y\sin a \] \[ y' = x\sin a + y\cos a \]

Problem 4

An airplane flies at a constant altitude \(h\) and speed 600 miles/hour. The angles of elevation change from \(20^\circ\) to \(60^\circ\) in one minute. Find the altitude.

Airplane angle of elevation

Solution

We first calculate distance \( d \) using the time and speed (1 minute = 1/60 hour) \[ d = 600 \times \frac{1}{60} = 10 \text{ miles} \] We next express the tangent of the given angles of elevation as follows \[ \tan(20^\circ) = \frac{h}{d+x},\quad \tan(60^\circ) = \frac{h}{x} \] Eliminating \(x\): \[ h = \frac{d}{\frac{1}{\tan(20^\circ)} - \frac{1}{\tan(60^\circ)}} = 4.60 \text{ miles} \]

Problem 5

From point A (2000 m above ground), the angle of depression to a mountain top is \(15^\circ\). From point B on the ground, the angle of elevation is \(10^\circ\). Find the height of the mountain.

Mountain height trigonometry problem

Solution

Mountain height trigonometry problem for solution to problem 5
From the first right triangle, we write: \[ \tan(10^\circ) = \frac{h}{d}. \] From the second right triangle, we write: \[ \tan(15^\circ) = \frac{2000 - h}{d}. \] Solve each equation for \(d\): \[ d = \frac{h}{\tan(10^\circ)} \quad \text{and} \quad d = \frac{2000 - h}{\tan(15^\circ)}. \] Eliminate \(d\) by equating the two expressions: \[ \frac{h}{\tan(10^\circ)} = \frac{2000 - h}{\tan(15^\circ)}. \] Solve for \(h\): \[ h = \frac{2000 \tan(10^\circ)} {\tan(15^\circ) + \tan(10^\circ)}. \] Evaluating numerically: \[ h \approx 793.8 \text{ m}. \]

More practice: Trigonometry Problems and Tutorials