In - ABC - AB -10 cm - - A -40- - B -110- - The area of the triangle will be equal to - T- an 40-0-84- T an 70-2-75A- 60-47 cm 2B- 68-19 cm 2C- 70-8 cm 2D- 50-15 cm 2

The correct option is A nbsp;60.47 cm^2 nbsp; nbsp; Draw CD perpendicular to AB extended to D. nbsp; nbsp; Let nbsp; BD=x CD=h nbsp; nbsp; ∠ ...

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Standard X

Mathematics

Heights and Distances

In Δ ABC , AB...

Question

In $Δ A B C, A B = 10 c m, ∠ A = 40^{\circ}, ∠ B = 110^{\circ} .$ The area of the triangle will be equal to ______. $[T a n 40^{\circ} = 0.84, T a n 70^{\circ} = 2.75]$

$70.8 c m^{2}$

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$60.47 c m^{2}$

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$50.15 c m^{2}$

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$68.19 c m^{2}$

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Solution

The correct option is B
$60.47 c m^{2}$

Draw CD perpendicular to AB extended to D.

Let $B D = x & C D = h$

$∠ C B D = (180 - 110^{\circ}) = 70^{\circ}$

In Triangle CBD

$T a n 70^{\circ} = \frac{C D}{B D} = \frac{h}{x}$

$h = x T a n 70^{\circ}$

$h = 2.75 x . . . (i)$

In triangle ACD

$T a n 40^{\circ} = \frac{C D}{A D} = \frac{h}{(10 + x)}$

$h = (10 + x) T a n 40^{\circ}$

$h = (10 + x) .0 .84$

$h = 8.4 + 0.84 x . . . (i i)$

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

$2.75 x = 8.4 + 0.84 x$

$1.91 x = 8.4$

$x = 4.39 c m$

$h = 2.75 x$ From (i)

$h = 2.75 \times 4.39$

$h = 12.09 c m$

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

$= \frac{1}{2} \times 10 \times 12.09$

$= 60.47 c m^{2}$

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