In - ABC - AB -15 cm - - A -30- - B -130- - The area of the triangle will be equal to - T an 30-0-57- T an 50-1-2 -A- 152-18 cm 2B- 122-13 cm 2C- 132-8 cm 2D- 142-12 cm 2

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Heights and Distances

In Δ ABC , AB...

Question

In $Δ A B C, A B = 15 c m, ∠ A = 30^{\circ}, ∠ B = 130^{\circ} .$ The area of the triangle will be equal to $[\begin{matrix} T a n 30^{\circ} = 0.57 T a n 50^{\circ} = 1.2 \end{matrix}]$

$132.8 c m^{2}$

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$152.18 c m^{2}$

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$122.13 c m^{2}$

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$142.12 c m^{2}$

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Solution

The correct option is C
$122.13 c m^{2}$

Draw CD perpendicular to AB extended till D.

$∠ C B D = 180 - 130^{\circ} = 50^{\circ}$

$T a n 50^{\circ} = \frac{C D}{B D} = \frac{h}{x}$

$h = x t a n 50^{\circ}$

$h = 1.2 x . . . (i)$

In $Δ A C D$

$T a n 30^{\circ} = \frac{C D}{A D} = \frac{h}{(15 + x)}$

$h = (15 + x) . T a n 30^{\circ}$

$h = (15 + x) .0 .57$

$h = 8.55 + 0.57 x -$ from ... (i)

From (i) $&$ (ii) we have

$1.2 x = 8.55 + 0.57 x$

$0.63 x = 8.55$

$x = 13.57$

$h = 1.2 x -$ from ...(i)

$h = 1.2 \times 13.57$

$h = 16.28$

Area of triangle ABC $= \frac{1}{2} \times A B \times h$

$= \frac{1}{2} \times 15 \times 16.28$

$= 122.13 c m^{2}$

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In Δ ABC, AB=15 cm, ∠ A=30∘, ∠ B=130∘. The area of the triangle will be equal to [Tan 30∘=0.57Tan 50∘=1.2]

In $Δ A B C, A B = 15 c m, ∠ A = 30^{\circ}, ∠ B = 130^{\circ} .$ The area of the triangle will be equal to $[\begin{matrix} T a n 30^{\circ} = 0.57 T a n 50^{\circ} = 1.2 \end{matrix}]$