Answer :
Sure! Let's solve the problem step-by-step.
### Given:
- [tex]\(AP = 12\)[/tex] meters
- [tex]\(GA = 5\)[/tex] meters
### Right-Angled Triangle [tex]\(GAP\)[/tex]:
The triangle is right-angled at [tex]\(A\)[/tex], so we apply the Pythagorean theorem to find [tex]\(GP\)[/tex].
[tex]\[ GA^2 + AP^2 = GP^2 \][/tex]
[tex]\[ 5^2 + 12^2 = GP^2 \][/tex]
[tex]\[ 25 + 144 = GP^2 \][/tex]
[tex]\[ GP^2 = 169 \][/tex]
[tex]\[ GP = \sqrt{169} \][/tex]
[tex]\[ GP = 13 \][/tex] meters
### 1. Finding [tex]\(\tan x\)[/tex]:
In [tex]\(\triangle AGT\)[/tex], we use the definition of the tangent function:
[tex]\[ \tan x = \frac{\text{opposite}}{\text{adjacent}} \][/tex]
Since [tex]\(GA\)[/tex] is opposite [tex]\(x\)[/tex] and [tex]\(AT\)[/tex] is adjacent:
[tex]\[ \tan x = \frac{GA}{AP} = \frac{5}{12} \][/tex]
### 2. Finding [tex]\(\tan(2x)\)[/tex] using the double angle formula:
The double angle formula for tangent is:
[tex]\[ \tan(2x) = \frac{2 \tan(x)}{1 - \tan^2(x)} \][/tex]
Let's substitute [tex]\(\tan x = \frac{5}{12}\)[/tex]:
[tex]\[ \tan(2x) = \frac{2 \cdot \frac{5}{12}}{1 - \left(\frac{5}{12}\right)^2} \][/tex]
[tex]\[ = \frac{\frac{10}{12}}{1 - \frac{25}{144}} \][/tex]
[tex]\[ = \frac{\frac{10}{12}}{\frac{144}{144} - \frac{25}{144}} \][/tex]
[tex]\[ = \frac{\frac{10}{12}}{\frac{119}{144}} \][/tex]
[tex]\[ = \frac{10 \cdot 144}{12 \cdot 119} \][/tex]
[tex]\[ = \frac{10 \cdot 12}{119} \][/tex]
[tex]\[ = \frac{120}{119} \][/tex]
Thus:
[tex]\[ \tan(2x) = \frac{120}{119} \][/tex]
### 3. Finding length [tex]\(AT\)[/tex]:
In [tex]\(\triangle AGT\)[/tex], we know [tex]\(\tan(x)\)[/tex]:
[tex]\[ \tan x = \frac{GA}{AT} \][/tex]
[tex]\[ \frac{5}{AT} = \frac{5}{12} \][/tex]
[tex]\[ AT = 12 \][/tex]
### Final Answers:
a. [tex]\(\tan(2x) = \frac{120}{119} = 1.0084\)[/tex] (approximately)
b. [tex]\(\tan x = \frac{5}{12} = 0.4167\)[/tex] (approximately)
c. [tex]\(AT = 12\)[/tex] meters
### Given:
- [tex]\(AP = 12\)[/tex] meters
- [tex]\(GA = 5\)[/tex] meters
### Right-Angled Triangle [tex]\(GAP\)[/tex]:
The triangle is right-angled at [tex]\(A\)[/tex], so we apply the Pythagorean theorem to find [tex]\(GP\)[/tex].
[tex]\[ GA^2 + AP^2 = GP^2 \][/tex]
[tex]\[ 5^2 + 12^2 = GP^2 \][/tex]
[tex]\[ 25 + 144 = GP^2 \][/tex]
[tex]\[ GP^2 = 169 \][/tex]
[tex]\[ GP = \sqrt{169} \][/tex]
[tex]\[ GP = 13 \][/tex] meters
### 1. Finding [tex]\(\tan x\)[/tex]:
In [tex]\(\triangle AGT\)[/tex], we use the definition of the tangent function:
[tex]\[ \tan x = \frac{\text{opposite}}{\text{adjacent}} \][/tex]
Since [tex]\(GA\)[/tex] is opposite [tex]\(x\)[/tex] and [tex]\(AT\)[/tex] is adjacent:
[tex]\[ \tan x = \frac{GA}{AP} = \frac{5}{12} \][/tex]
### 2. Finding [tex]\(\tan(2x)\)[/tex] using the double angle formula:
The double angle formula for tangent is:
[tex]\[ \tan(2x) = \frac{2 \tan(x)}{1 - \tan^2(x)} \][/tex]
Let's substitute [tex]\(\tan x = \frac{5}{12}\)[/tex]:
[tex]\[ \tan(2x) = \frac{2 \cdot \frac{5}{12}}{1 - \left(\frac{5}{12}\right)^2} \][/tex]
[tex]\[ = \frac{\frac{10}{12}}{1 - \frac{25}{144}} \][/tex]
[tex]\[ = \frac{\frac{10}{12}}{\frac{144}{144} - \frac{25}{144}} \][/tex]
[tex]\[ = \frac{\frac{10}{12}}{\frac{119}{144}} \][/tex]
[tex]\[ = \frac{10 \cdot 144}{12 \cdot 119} \][/tex]
[tex]\[ = \frac{10 \cdot 12}{119} \][/tex]
[tex]\[ = \frac{120}{119} \][/tex]
Thus:
[tex]\[ \tan(2x) = \frac{120}{119} \][/tex]
### 3. Finding length [tex]\(AT\)[/tex]:
In [tex]\(\triangle AGT\)[/tex], we know [tex]\(\tan(x)\)[/tex]:
[tex]\[ \tan x = \frac{GA}{AT} \][/tex]
[tex]\[ \frac{5}{AT} = \frac{5}{12} \][/tex]
[tex]\[ AT = 12 \][/tex]
### Final Answers:
a. [tex]\(\tan(2x) = \frac{120}{119} = 1.0084\)[/tex] (approximately)
b. [tex]\(\tan x = \frac{5}{12} = 0.4167\)[/tex] (approximately)
c. [tex]\(AT = 12\)[/tex] meters