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L.H.S

sin⁑x=14=ph\sin x=\frac{1}{4}=\frac{p}{h}

b=(4)2βˆ’(1)2b=\sqrt[]{\left(4\right)^2-\left(1\right)^2}

=15=\sqrt[]{15}

Here h= 4, p=1. To get the value of 'b' we use b=h2βˆ’p2b=\sqrt[]{h^2-p^2}

Since 'x' lies in 2nd quadrant

cos⁑x=βˆ’154\cos x=-\frac{\sqrt[]{15}}{4}

... (More)

BCAβ€²C=tan⁑ 30∘\frac{BC}{A'C}=\tan \ 30^{\circ }

BC8=13\frac{BC}{8}=\frac{1}{\sqrt[]{3}}

BCΒ =Β 83BC\ =\ \frac{8}{\sqrt[]{3}}

Let AC be the original tree. It was broken into two parts.

Broken part is A'B and ΞΈ\theta is angle between the broken part A'B and grou... (More)

sin⁑θ=ABAC\sin \theta = \frac{AB}{AC}

Β 

Assume ACAC is the long rope and ABAB is the vertical pole. ΞΈ\theta is the angle between the rope and the ground.Β 

sin⁑θ=ABAC\sin \theta = \frac{AB}{AC}

sin⁑30o=AB20 m\sin 30^o = \frac{AB}{20\ m}
Substitute valu... (More)

cos⁑x=βˆ’13=bh\cos x=-\frac{1}{3}=\frac{b}{h}

p=(3)2βˆ’(1)2p=\sqrt[]{\left(3\right)^2-\left(1\right)^2}

=9βˆ’1=\sqrt[]{9-1}

=8=\sqrt[]{8}

=22=2\sqrt[]{2}

Here b=1b=1, h=3h= 3, then to get the value of bb we use p=h2βˆ’b2p=\sqrt[]{h^2-b^2}

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