Answer :
Let's solve each part step by step.
(a) Show that [tex]\(\angle AQB = 141^\circ\)[/tex].
Given:
- [tex]\(A\)[/tex] is due east of [tex]\(Q\)[/tex].
- The bearing of [tex]\(Q\)[/tex] from [tex]\(B\)[/tex] is [tex]\(051^\circ\)[/tex] (True Bearing).
1. Since [tex]\(A\)[/tex] is due east of [tex]\(Q\)[/tex], the bearing of [tex]\(A\)[/tex] from [tex]\(Q\)[/tex] is [tex]\(090^\circ\)[/tex] (due east).
2. Bearing of [tex]\(Q\)[/tex] from [tex]\(B\)[/tex] is [tex]\(051^\circ\)[/tex]. Therefore, the bearing of [tex]\(B\)[/tex] from [tex]\(Q\)[/tex] will be [tex]\(051^\circ + 180^\circ = 231^\circ\)[/tex] (since bearings are measured clockwise from the north).
Now, [tex]\( \angle AQB = 231^\circ - 90^\circ = 141^\circ \)[/tex].
Therefore, we have shown that [tex]\(\angle AQB = 141^\circ\)[/tex].
(b) Show that [tex]\(AQ = h \tan 78^\circ\)[/tex].
Given:
- The angle of elevation from [tex]\(A\)[/tex] to the top of the tower [tex]\(P\)[/tex] is [tex]\(12^\circ\)[/tex].
- [tex]\(h\)[/tex] is the height of the tower [tex]\(P\)[/tex].
We use the tangent of the angle of elevation:
[tex]\[ \tan(12^\circ) = \frac{h}{AQ} \][/tex]
Rearranging to solve for [tex]\(AQ\)[/tex]:
[tex]\[ AQ = \frac{h}{\tan(12^\circ)} \][/tex]
Using cotangent ([tex]\(\cot\)[/tex]), we have:
[tex]\[ AQ = h \cot(12^\circ) \][/tex]
Since [tex]\(\cot(12^\circ)\)[/tex] is the same as [tex]\(\tan(78^\circ)\)[/tex] because [tex]\(\tan(78^\circ) = \frac{1}{\cot(78^\circ)} = \cot(12^\circ)\)[/tex], we get:
[tex]\[ AQ = h \tan(78^\circ) \][/tex]
Therefore, we have shown that [tex]\(AQ = h \tan(78^\circ)\)[/tex].
(c) Show that [tex]\(h = \frac{1000}{\sqrt{\tan^2 78 + \tan^2 79 - 2 \tan 78 \tan 79 \cos 141}}\)[/tex].
To prove this, let's first use the Law of Cosines in [tex]\(\triangle AQB\)[/tex].
Given:
- [tex]\( AB = 1000 \)[/tex] meters,
- [tex]\(\angle AQB = 141^\circ\)[/tex],
- [tex]\( AQ = h \tan 78^\circ\)[/tex],
- [tex]\( BQ = h \tan 79^\circ\)[/tex].
From the Law of Cosines:
[tex]\[ AB^2 = AQ^2 + BQ^2 - 2 \cdot AQ \cdot BQ \cdot \cos (\angle AQB) \][/tex]
Substitute the given values:
[tex]\[ 1000^2 = (h \tan 78^\circ)^2 + (h \tan 79^\circ)^2 - 2 \cdot (h \tan 78^\circ) \cdot (h \tan 79^\circ) \cdot \cos 141^\circ \][/tex]
Simplify:
[tex]\[ 1000^2 = h^2 (\tan^2 78^\circ + \tan^2 79^\circ - 2 \cdot \tan 78^\circ \cdot \tan 79^\circ \cdot \cos 141^\circ) \][/tex]
Solving for [tex]\(h^2\)[/tex]:
[tex]\[ h^2 = \frac{1000^2}{\tan^2 78^\circ + \tan^2 79^\circ - 2 \cdot \tan 78^\circ \cdot \tan 79^\circ \cdot \cos 141^\circ} \][/tex]
Therefore:
[tex]\[ h = \frac{1000}{\sqrt{\tan^2 78^\circ + \tan^2 79^\circ - 2 \cdot \tan 78^\circ \cdot \tan 79^\circ \cdot \cos 141^\circ}} \][/tex]
Thus, we have confirmed that [tex]\(h = \frac{1000}{\sqrt{\tan^2 78 + \tan^2 79 - 2 \tan 78 \tan 79 \cos 141}}\)[/tex].
(a) Show that [tex]\(\angle AQB = 141^\circ\)[/tex].
Given:
- [tex]\(A\)[/tex] is due east of [tex]\(Q\)[/tex].
- The bearing of [tex]\(Q\)[/tex] from [tex]\(B\)[/tex] is [tex]\(051^\circ\)[/tex] (True Bearing).
1. Since [tex]\(A\)[/tex] is due east of [tex]\(Q\)[/tex], the bearing of [tex]\(A\)[/tex] from [tex]\(Q\)[/tex] is [tex]\(090^\circ\)[/tex] (due east).
2. Bearing of [tex]\(Q\)[/tex] from [tex]\(B\)[/tex] is [tex]\(051^\circ\)[/tex]. Therefore, the bearing of [tex]\(B\)[/tex] from [tex]\(Q\)[/tex] will be [tex]\(051^\circ + 180^\circ = 231^\circ\)[/tex] (since bearings are measured clockwise from the north).
Now, [tex]\( \angle AQB = 231^\circ - 90^\circ = 141^\circ \)[/tex].
Therefore, we have shown that [tex]\(\angle AQB = 141^\circ\)[/tex].
(b) Show that [tex]\(AQ = h \tan 78^\circ\)[/tex].
Given:
- The angle of elevation from [tex]\(A\)[/tex] to the top of the tower [tex]\(P\)[/tex] is [tex]\(12^\circ\)[/tex].
- [tex]\(h\)[/tex] is the height of the tower [tex]\(P\)[/tex].
We use the tangent of the angle of elevation:
[tex]\[ \tan(12^\circ) = \frac{h}{AQ} \][/tex]
Rearranging to solve for [tex]\(AQ\)[/tex]:
[tex]\[ AQ = \frac{h}{\tan(12^\circ)} \][/tex]
Using cotangent ([tex]\(\cot\)[/tex]), we have:
[tex]\[ AQ = h \cot(12^\circ) \][/tex]
Since [tex]\(\cot(12^\circ)\)[/tex] is the same as [tex]\(\tan(78^\circ)\)[/tex] because [tex]\(\tan(78^\circ) = \frac{1}{\cot(78^\circ)} = \cot(12^\circ)\)[/tex], we get:
[tex]\[ AQ = h \tan(78^\circ) \][/tex]
Therefore, we have shown that [tex]\(AQ = h \tan(78^\circ)\)[/tex].
(c) Show that [tex]\(h = \frac{1000}{\sqrt{\tan^2 78 + \tan^2 79 - 2 \tan 78 \tan 79 \cos 141}}\)[/tex].
To prove this, let's first use the Law of Cosines in [tex]\(\triangle AQB\)[/tex].
Given:
- [tex]\( AB = 1000 \)[/tex] meters,
- [tex]\(\angle AQB = 141^\circ\)[/tex],
- [tex]\( AQ = h \tan 78^\circ\)[/tex],
- [tex]\( BQ = h \tan 79^\circ\)[/tex].
From the Law of Cosines:
[tex]\[ AB^2 = AQ^2 + BQ^2 - 2 \cdot AQ \cdot BQ \cdot \cos (\angle AQB) \][/tex]
Substitute the given values:
[tex]\[ 1000^2 = (h \tan 78^\circ)^2 + (h \tan 79^\circ)^2 - 2 \cdot (h \tan 78^\circ) \cdot (h \tan 79^\circ) \cdot \cos 141^\circ \][/tex]
Simplify:
[tex]\[ 1000^2 = h^2 (\tan^2 78^\circ + \tan^2 79^\circ - 2 \cdot \tan 78^\circ \cdot \tan 79^\circ \cdot \cos 141^\circ) \][/tex]
Solving for [tex]\(h^2\)[/tex]:
[tex]\[ h^2 = \frac{1000^2}{\tan^2 78^\circ + \tan^2 79^\circ - 2 \cdot \tan 78^\circ \cdot \tan 79^\circ \cdot \cos 141^\circ} \][/tex]
Therefore:
[tex]\[ h = \frac{1000}{\sqrt{\tan^2 78^\circ + \tan^2 79^\circ - 2 \cdot \tan 78^\circ \cdot \tan 79^\circ \cdot \cos 141^\circ}} \][/tex]
Thus, we have confirmed that [tex]\(h = \frac{1000}{\sqrt{\tan^2 78 + \tan^2 79 - 2 \tan 78 \tan 79 \cos 141}}\)[/tex].
