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### Q2. If [tex]\(12 \cot \theta = 15\)[/tex], then find [tex]\(\sec \theta\)[/tex]:
1. Given [tex]\(12 \cot \theta = 15\)[/tex], we can solve for [tex]\(\cot \theta\)[/tex]:
[tex]\[ \cot \theta = \frac{15}{12} = 1.25 \][/tex]
2. Since [tex]\(\cot \theta = \frac{1}{\tan \theta}\)[/tex], we find:
[tex]\[ \tan \theta = \frac{1}{1.25} = \frac{4}{5} = 0.8 \][/tex]
3. We know that [tex]\(\sec^2 \theta = 1 + \tan^2 \theta\)[/tex]:
[tex]\[ \sec^2 \theta = 1 + (0.8)^2 = 1 + 0.64 = 1.64 \][/tex]
4. Taking the square root to find [tex]\(\sec \theta\)[/tex]:
[tex]\[ \sec \theta = \sqrt{1.64} \approx 1.2806 \][/tex]
So, [tex]\(\sec \theta \approx 1.2806\)[/tex].
### Q3. In [tex]\(\triangle PQR\)[/tex], right-angled at [tex]\(Q\)[/tex], [tex]\(PR + QR = 30 \text{ cm}\)[/tex] and [tex]\(PQ = 10 \text{ cm}\)[/tex]. Determine the values of [tex]\(\sin P\)[/tex], [tex]\(\cos P\)[/tex], and [tex]\(\tan P\)[/tex]:
1. Let's assign the following:
- [tex]\(PQ = 10 \text{ cm}\)[/tex]
- Let [tex]\(PR = x \text{ cm}\)[/tex]
- Then [tex]\(QR = 30 - x \text{ cm}\)[/tex]
2. Using the Pythagorean theorem for the right-angled triangle:
[tex]\[ PR^2 = PQ^2 + QR^2 \][/tex]
Substituting the values:
[tex]\[ x^2 = 10^2 + (30 - x)^2 \][/tex]
3. Expanding and simplifying:
[tex]\[ x^2 = 100 + 900 - 60x + x^2 \][/tex]
[tex]\[ 0 = 1000 - 60x \][/tex]
[tex]\[ 60x = 1000 \][/tex]
[tex]\[ x \approx 16.6667 \text{ cm} \][/tex]
Therefore, [tex]\(PR \approx 16.6667 \text{ cm}\)[/tex] and [tex]\(QR \approx 30 - 16.6667 = 13.3333 \text{ cm}\)[/tex].
4. Now, [tex]\(\sin P, \cos P\)[/tex], and [tex]\(\tan P\)[/tex] can be found:
[tex]\[ \sin P = \frac{opposite}{hypotenuse} = \frac{QR}{PR} \approx \frac{13.3333}{16.6667} \approx 0.8 \][/tex]
[tex]\[ \cos P = \frac{adjacent}{hypotenuse} = \frac{PQ}{PR} \approx \frac{10}{16.6667} \approx 0.6 \][/tex]
[tex]\[ \tan P = \frac{opposite}{adjacent} = \frac{QR}{PQ} \approx \frac{13.3333}{10} \approx 1.3333 \][/tex]
So, the values are:
- [tex]\(\sin P \approx 0.8\)[/tex]
- [tex]\(\cos P \approx 0.6\)[/tex]
- [tex]\(\tan P \approx 1.3333\)[/tex]
### Q4. If [tex]\(\sec 4 \theta = \csc (\theta - 300)\)[/tex], where [tex]\(4 \theta\)[/tex] is an acute angle, find the value of [tex]\(A\)[/tex]:
1. Given the relationship [tex]\(\sec 4 \theta = \csc (\theta - 300)\)[/tex], rewrite using trigonometric identities:
[tex]\[ \sec 4 \theta = \frac{1}{\cos 4 \theta} \quad \text{and} \quad \csc (\theta - 300) = \frac{1}{\sin (\theta - 300)} \][/tex]
Thus:
[tex]\[ \frac{1}{\cos 4 \theta} = \frac{1}{\sin (\theta - 300)} \][/tex]
Therefore:
[tex]\[ \cos 4 \theta = \sin (\theta - 300) \][/tex]
2. Using the trigonometric identity [tex]\(\cos x = \sin (90^\circ - x)\)[/tex]:
[tex]\[ \cos 4 \theta = \sin (\theta - 300) \implies 4 \theta = 90^\circ - (\theta - 300) \][/tex]
Solving for [tex]\(\theta\)[/tex]:
[tex]\[ 4 \theta = 390^\circ - \theta \][/tex]
[tex]\[ 5 \theta = 390^\circ \][/tex]
[tex]\[ \theta = 78^\circ \][/tex]
Therefore, the value of [tex]\(A\)[/tex] is [tex]\(78^\circ\)[/tex].
### Q2. If [tex]\(12 \cot \theta = 15\)[/tex], then find [tex]\(\sec \theta\)[/tex]:
1. Given [tex]\(12 \cot \theta = 15\)[/tex], we can solve for [tex]\(\cot \theta\)[/tex]:
[tex]\[ \cot \theta = \frac{15}{12} = 1.25 \][/tex]
2. Since [tex]\(\cot \theta = \frac{1}{\tan \theta}\)[/tex], we find:
[tex]\[ \tan \theta = \frac{1}{1.25} = \frac{4}{5} = 0.8 \][/tex]
3. We know that [tex]\(\sec^2 \theta = 1 + \tan^2 \theta\)[/tex]:
[tex]\[ \sec^2 \theta = 1 + (0.8)^2 = 1 + 0.64 = 1.64 \][/tex]
4. Taking the square root to find [tex]\(\sec \theta\)[/tex]:
[tex]\[ \sec \theta = \sqrt{1.64} \approx 1.2806 \][/tex]
So, [tex]\(\sec \theta \approx 1.2806\)[/tex].
### Q3. In [tex]\(\triangle PQR\)[/tex], right-angled at [tex]\(Q\)[/tex], [tex]\(PR + QR = 30 \text{ cm}\)[/tex] and [tex]\(PQ = 10 \text{ cm}\)[/tex]. Determine the values of [tex]\(\sin P\)[/tex], [tex]\(\cos P\)[/tex], and [tex]\(\tan P\)[/tex]:
1. Let's assign the following:
- [tex]\(PQ = 10 \text{ cm}\)[/tex]
- Let [tex]\(PR = x \text{ cm}\)[/tex]
- Then [tex]\(QR = 30 - x \text{ cm}\)[/tex]
2. Using the Pythagorean theorem for the right-angled triangle:
[tex]\[ PR^2 = PQ^2 + QR^2 \][/tex]
Substituting the values:
[tex]\[ x^2 = 10^2 + (30 - x)^2 \][/tex]
3. Expanding and simplifying:
[tex]\[ x^2 = 100 + 900 - 60x + x^2 \][/tex]
[tex]\[ 0 = 1000 - 60x \][/tex]
[tex]\[ 60x = 1000 \][/tex]
[tex]\[ x \approx 16.6667 \text{ cm} \][/tex]
Therefore, [tex]\(PR \approx 16.6667 \text{ cm}\)[/tex] and [tex]\(QR \approx 30 - 16.6667 = 13.3333 \text{ cm}\)[/tex].
4. Now, [tex]\(\sin P, \cos P\)[/tex], and [tex]\(\tan P\)[/tex] can be found:
[tex]\[ \sin P = \frac{opposite}{hypotenuse} = \frac{QR}{PR} \approx \frac{13.3333}{16.6667} \approx 0.8 \][/tex]
[tex]\[ \cos P = \frac{adjacent}{hypotenuse} = \frac{PQ}{PR} \approx \frac{10}{16.6667} \approx 0.6 \][/tex]
[tex]\[ \tan P = \frac{opposite}{adjacent} = \frac{QR}{PQ} \approx \frac{13.3333}{10} \approx 1.3333 \][/tex]
So, the values are:
- [tex]\(\sin P \approx 0.8\)[/tex]
- [tex]\(\cos P \approx 0.6\)[/tex]
- [tex]\(\tan P \approx 1.3333\)[/tex]
### Q4. If [tex]\(\sec 4 \theta = \csc (\theta - 300)\)[/tex], where [tex]\(4 \theta\)[/tex] is an acute angle, find the value of [tex]\(A\)[/tex]:
1. Given the relationship [tex]\(\sec 4 \theta = \csc (\theta - 300)\)[/tex], rewrite using trigonometric identities:
[tex]\[ \sec 4 \theta = \frac{1}{\cos 4 \theta} \quad \text{and} \quad \csc (\theta - 300) = \frac{1}{\sin (\theta - 300)} \][/tex]
Thus:
[tex]\[ \frac{1}{\cos 4 \theta} = \frac{1}{\sin (\theta - 300)} \][/tex]
Therefore:
[tex]\[ \cos 4 \theta = \sin (\theta - 300) \][/tex]
2. Using the trigonometric identity [tex]\(\cos x = \sin (90^\circ - x)\)[/tex]:
[tex]\[ \cos 4 \theta = \sin (\theta - 300) \implies 4 \theta = 90^\circ - (\theta - 300) \][/tex]
Solving for [tex]\(\theta\)[/tex]:
[tex]\[ 4 \theta = 390^\circ - \theta \][/tex]
[tex]\[ 5 \theta = 390^\circ \][/tex]
[tex]\[ \theta = 78^\circ \][/tex]
Therefore, the value of [tex]\(A\)[/tex] is [tex]\(78^\circ\)[/tex].
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