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Reducing a Triangle to Two Right Angled Triangles

In the given ...

Question

In the given triangle $A B C, A B = 8 c m, ∠ A = 50^{\circ}, ∠ B = 70^{\circ}$ as shown in the figure. The perpendicular distance from C to AB is equal to $[t a n 50^{\circ} = 1.2, t a n 70^{\circ} = 2.75]$

10.6 cm

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8.6 cm

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6.69 cm

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7.6 cm

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Solution

The correct option is C
6.69 cm

Draw CD perpendicular to AB

In triangle ADC

$t a n 50^{\circ} = \frac{C D}{A D} = \frac{h}{x}$

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

$h = 1.2 x . . (i)$

In triangle CBD

$h = (8 - x) t a n 70^{\circ}$

$h = (8 - x) 2.74$

$h = 21.92 - 2.74 x . . . (i i)$

from (i) and (ii) we get

$1.2 x = 21.92 - 2.74 x$

$3.94 x = 21.92$

$x = 5.57 c m$

$h = 1.2 x$

$h = 1.2 \times 5.56 = 6.68 c m$

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In the given triangle ABC, AB=8cm, ∠A=50∘, ∠B=70∘ as shown in the figure. The perpendicular distance from C to AB is equal to [tan 50∘=1.2,tan 70∘=2.75]

The correct option is C 6.69 cm Draw CD perpendicular to AB In triangle ADC tan 50∘=CDAD=hx h=x tan 50∘ h=1.2x ..(i) In triangle CBD h=(8−x)tan 70∘ h=(8−x)2.74 h=21.92−2.74 x ...(ii) from (i) and (ii) we get 1.2x=21.92−2.74 x 3.94 x=21.92 x=5.57cm h=1.2x h=1.2×5.56=6.68 cm

In the given triangle $A B C, A B = 8 c m, ∠ A = 50^{\circ}, ∠ B = 70^{\circ}$ as shown in the figure. The perpendicular distance from C to AB is equal to $[t a n 50^{\circ} = 1.2, t a n 70^{\circ} = 2.75]$