Question

Please answer all of these ​

Answer 1

${\bold{\underline{\underline{Step-by-step\:explanation:}}}}$

• For 1st Question :

$\implies sin \theta = \frac{12}{13} = \frac{p}{b} \\ \\ \bold{By \: Phythagoras : } \\ \implies {h}^{2} = {p}^{2} + {b}^{2} \\ \\ \implies ({13})^{2} = ({12})^{2} + {b}^{2} \\ \\ \implies {b}^{2} = 25 \\ \\ \implies b = 5 \\ \\ \bold{for \: cos \theta : } \\ \bold{\implies cos \theta = \frac{b}{h} = \frac{5}{13} }$

• For 2nd Question :

$\implies sin \: 45 \degree - cos \: 45 \degree \\ \\ \implies \frac{1}{\sqrt{2} } - \frac{1}{ \sqrt{2} } \\ \\ \implies 0$

• For 3rd Question :

$\implies {sin}^{2} \:48 \degree + {cos}^{2} \: 48 \degree \\ \\ \bold{ \implies{sin}^{2} \: \theta + {cos}^{2} \: \theta = 1} \\ \\ \implies 1$

• For 4th Question :

$\implies sin(45 \degree + \theta) - cos(45 \degree - \theta) \\ \\ \implies sin(45 \degree + \theta) - cos(90 \degree - 45 \degree + \theta) \\ \\ \bold{ \implies cos(90 - \theta) = sin \theta} \\ \\ \implies sin(45 \degree + \theta) - sin(45 \degree + \theta) \\ \\ \implies 0$

• For 5th Question :

$\implies tan \: \alpha = \sqrt{3} \\ \\ \bold{\implies \alpha = 60 \degree }\\ \\ \implies tan \: \beta = \frac{1}{ \sqrt{3} } \\ \\ \bold{\implies \beta = 30 \degree} \\ \\ \bold{For \: Finding \: value : } \\ \implies cot( \alpha + \beta ) \\ \\ \implies cot(60 \degree + 30 \degree) \\ \\ \bold{\implies cot \: 90 \degree = 0 }$

• For 6th Question :

$\implies \frac{tan \: 30 \degree}{cot \: 60 \degree} \\ \\ \implies \frac{ \frac{1}{ \sqrt{3} } }{ \frac{1}{ \sqrt{3} } } \\ \\ \implies \frac{ \sqrt{3} }{ \sqrt{3} } \\ \\ \implies 1$

• For 7th Question :

$\implies \angle c = 90 \degree \\ \\ \implies \angle a + \angle b = 90 \degree \\ \\ \bold{For \: Finding \: value } \\ \implies cos(a + b) \\ \\ \implies cos \:90 \degree \\ \\ \implies 1$

• For 8th Question :

$\implies sin \:a = \frac{1}{2} \\ \\ \bold{\implies a = 30 \degree} \\ \\ \bold{for \: cos \: a} \\ \implies cos \: a \\ \\ \implies \bold{cos \: 30 \degree = \frac{ \sqrt{3} }{2} }$

• For 9th Question :

$\implies tan \: \theta = \frac{p}{b} \\ \\ \implies tan \: \theta = \frac{6}{2 \sqrt{3} } \\ \\ \implies tan \: \theta = \frac{3}{ \sqrt{3} } \\ \\ \implies tan \: \theta = \sqrt{3} \\ \\ \implies \bold{\theta = 60 \degree}$