L.H.S $\sin x=\frac{1}{4}=\frac{p}{h}$ $b=\sqrt[]{\left(4\right)^2-\left(1\right)^2}$ $=\sqrt[]{15}$ | Here h= 4, p=1. To get the value of 'b' we use $b=\sqrt[]{h^2-p^2}$ |

Since 'x' lies in 2nd quadrant $\cos x=-\frac{\sqrt[]{15}}{4}$ ... (More) |

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AnswerMathTrigonometry

AnswerMathTrigonometry

$\frac{BC}{A'C}=\tan \ 30^{\circ }$ $\frac{BC}{8}=\frac{1}{\sqrt[]{3}}$ $BC\ =\ \frac{8}{\sqrt[]{3}}$ | Let AC be the original tree. It was broken into two parts. Broken part is A'B and $\theta$ is angle between the broken part A'B and grou... (More) |

AnswerMathTrigonometry

$\sin \theta = \frac{AB}{AC}$ Β | Assume $AC$ is the long rope and $AB$ is the vertical pole. $\theta$ is the angle between the rope and the ground.Β $\sin \theta = \frac{AB}{AC}$ |

$\sin 30^o = \frac{AB}{20\ m}$ | Substitute valu... (More) |

AnswerMathTrigonometry

$\cos x=-\frac{1}{3}=\frac{b}{h}$ $p=\sqrt[]{\left(3\right)^2-\left(1\right)^2}$ $=\sqrt[]{9-1}$ $=\sqrt[]{8}$ $=2\sqrt[]{2}$ | Here $b=1$, $h= 3$, then to get the value of $b$ we use $p=\sqrt[]{h^2-b^2}$ |

Since 'x' lies in quadr... (More) |