{"id":246286,"date":"2025-03-31T16:00:40","date_gmt":"2025-03-31T08:00:40","guid":{"rendered":"https:\/\/thepiquelab.com\/blog\/?p=246286"},"modified":"2025-08-12T15:20:47","modified_gmt":"2025-08-12T07:20:47","slug":"trigonometry-how-to-solve-bearings-questions","status":"publish","type":"post","link":"https:\/\/thepiquelab.com\/blog\/trigonometry-how-to-solve-bearings-questions\/","title":{"rendered":"Trigonometry: How To Solve Bearings Questions"},"content":{"rendered":"

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Introduction<\/strong><\/p>\n

Welcome to a new blog post on Trigonometry<\/strong>!<\/p>\n

In this article, we will be analysing a question on the topic of Trigonometry<\/strong> from the 2020 Punggol Secondary School (PGSS) S3E SA2 Examination Paper<\/strong>.<\/p>\n

Read this blog post if you want to solve Math questions involving Bearings better and faster.<\/p>\n

Alternatively, you can also watch my explainer video<\/a> for free by visiting The Pique Lab\u2019s YouTube channel<\/a>!<\/p>\n

\"Click<\/a><\/p>\n

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 <\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n

<\/p>\n\n

Let’s Take A Look At This Trigonometry Question<\/span><\/h2>\n

<\/p>\n

<\/p>\n

\"The<\/p>\n

Source: Punggol Secondary School \u2014 2020 S3E SA2 Examination Paper [Q7]<\/span><\/em>
<\/p>\n

\n

Read Also:<\/h3>\n
    \n
  1. [Secondary 3 E-Math Explainer Video] Trigonometry : What Is The Bearing Of A From B?<\/b><\/a><\/li>\n
  2. Direct Proportion: Finding The Relationship Between Variables\u00a0<\/strong><\/a><\/li>\n<\/ol>\n<\/div>\n

    In this question, we are given a diagram, ABC<\/em>, which represents a horizontal triangular field and a vertical flagpole, AP<\/em>.<\/p>\n

    We are told that B<\/em> is 85 metres from A<\/em> on a bearing of 025\u00b0 and that point C<\/em> is 170 metres away from point A<\/em>.<\/p>\n

    We are also given length BC<\/em> as 180 metres and the height of the flagpole is 12 metres.<\/p>\n

    Let us start by establishing what a bearing is.<\/p>\n

    \n

    \ud83d\udd01\u00a0What Is A Bearing?<\/strong> \ud83d\udd01<\/p>\n

    A bearing is the angle in degrees<\/strong> which is measured clockwise from the north<\/strong> direction. It is usually stated as a three-digit system<\/strong>.<\/p>\n<\/blockquote>\n

    Let\u2019s say we have two points, A<\/em> and B<\/em>, and they are separated by a certain distance apart.<\/p>\n

    If we are interested in finding out the bearing of A<\/em> from B<\/em>, we must first identify the starting point.<\/p>\n

    \n

    \ud83c\udfc1\u00a0How To Identify The Starting Point<\/strong> \ud83c\udfc1<\/p>\n

    Remember that the starting point is always after the word from<\/strong>.<\/p>\n<\/blockquote>\n

    In (a)(i), B<\/em> is our start point and A<\/em> is our endpoint.<\/p>\n

    \"The<\/p>\n

    Source: Punggol Secondary School \u2014 2020 S3E SA2 Examination Paper [Q7]<\/span><\/em><\/p>\n

    <\/p>\n

    We shall now go to the start point and draw a north direction, which is typically drawn in the vertical direction as shown in the diagram below.<\/p>\n

    \"The<\/p>\n

    Remember that we want to find the bearing of A<\/span> from B<\/span> so we will draw a clockwise angle starting from the north direction and stop when it touches the green-coloured line.<\/p>\n

    \"The<\/p>\n

    Let\u2019s say that this angle is 10\u00b0. We\u2019ll say that the bearing of A<\/span> from B<\/span> is 010\u00b0<\/span>.<\/p>\n

    \"The<\/p>\n

    This is what it means by a three-digit system<\/strong>.<\/p>\n

    Now that we know what a bearing is, let us proceed to solve the question.<\/p>\n

    <\/p>\n

    Let’s Analyse Part (A)(I)<\/span><\/h2>\n

    \"The<\/p>\n

    Source: Punggol Secondary School \u2014 2020 S3E SA2 Examination Paper [Q7]<\/span><\/em><\/p>\n

    Since we are interested in finding the bearing of A<\/em> from B<\/em>, we must first find the start and end points.<\/p>\n

    Recall that the start point is always after the word from. In (a)(i), A<\/em> is our endpoint, while B<\/em> is our starting point.<\/p>\n

    In our diagram, we shall now proceed to point B and draw in the north direction.<\/p>\n

    \"How<\/p>\n

    Source: Punggol Secondary School \u2014 2020 S3E SA2 Examination Paper [Q7]<\/span><\/em><\/p>\n

    What do you notice in the north direction in our diagram?<\/p>\n

    The north is not pointing vertically upwards but rather is in a slanted direction.<\/p>\n

    \"How<\/p>\n

    Source: Punggol Secondary School \u2014 2020 S3E SA2 Examination Paper [Q7]<\/span><\/em><\/p>\n

    Remember to be mindful of that and draw in the north also a bit slanted.<\/p>\n

    \"How Source: Punggol Secondary School \u2014 2020 S3E SA2 Examination Paper [Q7]<\/span><\/em><\/p>\n

    So how do we calculate the bearing of A<\/span> from B<\/span>? Let us draw in the angle. It will be moving clockwise from the north, all the way until it touches the line connecting A<\/span> and B<\/span>.<\/p>\n

    \"How<\/p>\n

    Source: Punggol Secondary School \u2014 2020 S3E SA2 Examination Paper [Q7]<\/span><\/em><\/p>\n

    We are interested in finding the pink-coloured angle, but we will have to make use of the green-coloured angle below first.<\/p>\n

    \"How<\/p>\n

    Source: Punggol Secondary School \u2014 2020 S3E SA2 Examination Paper [Q7]<\/span><\/em><\/p>\n

    Notice that in our diagram, there are a total of two north directions drawn for us.<\/p>\n

    Because they are pointing in the same direction, we can conclude that these two are parallel to one another.<\/p>\n

    What deduction can we make about the 25\u00b0 we highlighted in blue and the green-coloured angle?<\/p>\n

    \"How<\/p>\n

    Source: Punggol Secondary School \u2014 2020 S3E SA2 Examination Paper [Q7]<\/span><\/em><\/p>\n

    Since they are created by two parallel lines, they are a pair of interior angles.<\/p>\n

    \n

    \ud83c\udf70\u00a0What Are Interior Angles? \ud83c\udf70<\/strong><\/p>\n

    They are two angles that sum up to 180\u00b0 and are formed in between parallel lines.<\/p>\n<\/blockquote>\n

    So how do we calculate the green-coloured angle?<\/p>\n

    We will take 180\u00b0 and subtract 25\u00b0 to get 155\u00b0.<\/p>\n

    <\/p>\n

    \"How<\/p>\n

     <\/p>\n

    Source: Punggol Secondary School \u2014 2020 S3E SA2 Examination Paper [Q7]<\/span><\/em><\/p>\n

    We can now proceed to solve part (a)(i) by taking one full revolution, which gives us 360\u00b0 around point B. Then, subtract the green-coloured angle of 155\u00b0 to get the final answer of 205\u00b0 as the bearing of A from B.<\/p>\n

    \"How<\/p>\n

    Source: Punggol Secondary School \u2014 2020 S3E SA2 Examination Paper [Q7]<\/span><\/em><\/p>\n

    Suggested Answer For Part (A)(I)<\/span><\/h2>\n

    205\u00b0<\/span><\/strong><\/p>\n

    <\/p>\n

    <\/p>\n

    <\/p>\n

    <\/p>\n

    <\/p>\n

    <\/p>\n

    <\/p>\n

    <\/p>\n

    <\/p>\n

    <\/p>\n

    <\/p>\n

    <\/p>\n

    <\/p>\n

    <\/p>\n

    <\/p>\n

    <\/h2>\n

    Let’s Analyse Part (A)(II)<\/span><\/h2>\n

    \"How<\/p>\n

    Source: Punggol Secondary School \u2014 2020 S3E SA2 Examination Paper [Q7]<\/span><\/em><\/p>\n

    In part (a)(ii), we are required to find \u2220<\/strong>ABC.<\/em><\/p>\n

    Let us start by drawing a simple sketch of the triangular plane ABC<\/em>.<\/p>\n

    \u2220<\/strong>ABC<\/em> is annotated by the green-coloured angle. How are we going to find its value?<\/p>\n

    \"How<\/p>\n

    Notice that we are given all three sides of the triangle, but we do not have any interior angles.<\/p>\n

    Hence, we will be using the Cosine Rule<\/strong> for this question.<\/p>\n

    \n

    \ud83d\udca1\u00a0What Is The Cosine Rule? \ud83d\udca1<\/strong><\/p>\n

    Cosine Rule helps us find unknown lengths and angles along a non-right angle triangle<\/strong>.<\/p>\n<\/blockquote>\n

    \"The<\/p>\n

    In this case, because we are interested in finding an unknown angle in a non-right angle triangle, we\u2019ll use the variation on the right.<\/p>\n

    Let us substitute the values into the formula.<\/p>\n

    \"The<\/p>\n

    \"\"<\/p>\n

    When we use the Cosine Rule, it is important to note the corresponding angles and the value we subtract in the numerator.<\/p>\n

    How do we identify this pairing here?<\/p>\n

    Because we\u2019re interested in \u2220<\/strong>ABC<\/em>, if we reference our diagram on the left, take note that the A<\/em> value we subtract encircled in blue will always be the side directly opposite the angle.<\/p>\n

    \"How<\/p>\n

    So in this case, the value of a<\/em> is 170, b<\/em> is 85, and c<\/em> is 180.<\/p>\n

    Let us set up the Cosine Rule now.<\/p>\n

    \"How<\/p>\n

    To find angle ABC<\/span> we shall take a cosine inverse of 143\/408.<\/p>\n

    \"How<\/p>\n

    <\/p>\n

    <\/p>\n

    Suggested Answer For Part (A)(II)<\/span>
    <\/h2>\n

    69.5\u00b0<\/span><\/strong><\/p>\n

    Let’s Analyse Part (B)<\/span><\/h2>\n

    \"How<\/p>\n

    Source: Punggol Secondary School \u2014 2020 S3E SA2 Examination Paper [Q7]<\/span><\/em><\/p>\n

    In part (b), we are given a rate relating the cost of the plot of land to its area in square metres.<\/p>\n

    We want to calculate the cost of the land occupied by the triangular field ABC<\/em>.<\/p>\n

    Let us start by finding the area of this plot of land.<\/p>\n

    \"How<\/p>\n

    <\/p>\n

    How do we go about it?<\/p>\n

    The area of a triangle can be calculated using the formula 1\/2 x base x height.<\/p>\n

    But are we going to use this to answer this question? It is not wrong<\/strong> to use it, but it will be very tedious as it involves multiple steps.<\/p>\n

    To help you save time, we\u2019ll use this formula:<\/p>\n

    \"\"<\/p>\n

    A<\/em> and B<\/em> here represent two sides of the triangle, while C<\/em> is what we call the included angle that is required to calculate the area itself.<\/p>\n

    What does it mean to be an included angle?<\/p>\n

    If we refer to the triangle below, we see that we have one angle given to us, which is 69.483\u00b0 calculated from part (a)(ii).<\/p>\n

    <\/p>\n

    \"How<\/p>\n

    This angle is created by two sides of the triangle \u2014 AB<\/em> and BC<\/em>, with the values of 85 and 180, respectively.<\/p>\n

    So together, these two sides here create the \u2220<\/strong>ABC<\/em>, which is what we call the included angle.<\/p>\n

    \"How<\/p>\n

    <\/p>\n

    Let us now substitute the values into the area formula. <\/p>\n

    \"How \"How<\/p>\n

    Therefore, the area is 7164.747 square metres.<\/p>\n

    To solve for the cost, we will multiply the area by the rate given in the question, which is $70.<\/p>\n

    \"How<\/h2>\n

    Suggested Answer For Part (B)<\/span><\/h2>\n

    $501 532.29 (2 d.p.)<\/span><\/strong><\/p>\n

    Let’s Analyse Part (C)<\/span><\/h2>\n

    \"How<\/p>\n

    Source: Punggol Secondary School \u2014 2020 S3E SA2 Examination Paper [Q7]<\/span><\/em><\/p>\n

    In part (c), we are given a man who is walking along BC<\/em> as shown in the diagram below.<\/p>\n

    Also, recall that in the question, there was a flagpole at point A<\/em> denoted by AP<\/em>, with a length of 12 metres.<\/p>\n

    \"How<\/p>\n

    In part (c), we are interested in calculating the greatest angle of elevation<\/strong> of the top of the flagpole when viewed by the man as he travels from point B<\/em> to point C.<\/em><\/p>\n

    Let us begin solving this by finding where exactly along BC<\/em> will we get the greatest angle of elevation.<\/p>\n

    It is the shortest distance from the base of the flag pole, which is denoted by point A<\/em> in the diagram.<\/p>\n

    Where exactly is that point?<\/p>\n

    It will be the perpendicular distance from point A<\/em>, all the way to line BC.<\/em><\/p>\n

    Let us label this as point M<\/em>. It is at this point where the greatest angle of elevation will be obtained.<\/p>\n

    \"How<\/p>\n

    If we connect points P<\/span> and M<\/span> in a straight line, we\u2019ll obtain a triangle as highlighted in yellow.<\/p>\n

    \"How<\/p>\n

    Let us sketch out triangle PAM<\/span><\/em>.<\/p>\n

    \"How<\/p>\n

    Since the vertical flag pole is placed on horizontal ground, we note that the angle PAM<\/em> is a 90-degree angle.<\/p>\n

    Where is the greatest angle of elevation in this triangle?<\/p>\n

    It will be at \u2220<\/strong>AMP.<\/em><\/p>\n

    How do we find this value?<\/p>\n

    We can make use of the trigonometry concepts that we learned before to evaluate this value.<\/p>\n

    When given an unknown angle in a right-angled triangle, we need a minimum of two sides.<\/p>\n

    At this point, we have one side \u2014 AP<\/em>. However, we do not have sides PM<\/em> and AM.<\/em><\/p>\n

    To solve for the angle required, we need to find either PM<\/em> or AM<\/em>. Which side should find?<\/p>\n

    For this question, we are going to look for side AM<\/em> because we have enough information to solve for this value.<\/p>\n

    Let us review the diagram.<\/p>\n

    What can we infer about AM<\/em> with respect to triangle ABC<\/em>?<\/p>\n

    \"How<\/p>\n

    It is actually the perpendicular height of the triangle itself.<\/p>\n

    Recall from part (b) that we found the area of triangle ABC<\/em> which was 7164.747.<\/p>\n

    Since we have the perpendicular height identified, we can make use of our 1\/2 x b<\/em> x h<\/em> formula.<\/p>\n

    We know that height was given as side AM.<\/em> So what should we choose as the base? It will be side BC<\/em>.<\/p>\n

    Let us now substitute the values to find AM<\/em>.<\/p>\n

    \"How<\/p>\n

    Going back to our triangle PAM<\/em>,<\/span> we can now label AM<\/span><\/em> as 79.608.<\/p>\n

    \"How<\/p>\n

    With two sides of the triangle, we can now use trigonometry to solve for the angle required.<\/p>\n

    AP<\/em> plays the role of our opposite side, while AM<\/em> plays the role of our adjacent side.<\/p>\n

    What trigonometry function should we use to solve for the angle required?<\/p>\n

    We will use the Tangent Function to obtain the greatest angle of elevation:<\/p>\n

    \"How<\/p>\n

    <\/p>\n

    <\/p>\n

    <\/p>\n

    <\/p>\n

    Suggested Answer For Part (C)<\/span><\/h2>\n

    8.6\u00b0<\/strong><\/span><\/p>\n

    Conclusion<\/span><\/h2>\n

    I hope that after reading this S3 A-Math blog post, you have learned the following:<\/p>\n