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$\left( {\sin 50^\circ = 0.77,\cos 50^\circ = 0.64,\tan 50^\circ = 1.19} \right)$

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Hint:Area of the triangle is half multiplied by its perpendicular times base.

Pictorial representation of given problem is shown above

Let, AD is perpendicular to BC.

So, the area of the triangle is half multiply by its perpendicular times base

$ = \dfrac{1}{2}\left( {AD} \right)\left( {BC} \right) = \dfrac{1}{2}\left( {AD} \right)\left( 6 \right) = 3AD$

Now in triangle ABD

$

\sin 50^\circ = \dfrac{{AD}}{{AB}} = \dfrac{{AD}}{5} \\

\Rightarrow AD = 5\sin 50^\circ = 5 \times 0.77 = 3.85cm \\

$

Therefore area of triangle is

Area$ = 3AD = 3 \times 3.85 = 11.55c{m^2}$

Now in triangle ABD

$

\tan 50^\circ = \dfrac{{AD}}{{BD}} = \dfrac{{3.85}}{{BD}} \\

\Rightarrow BD = \dfrac{{3.85}}{{\tan 50^\circ }} = \dfrac{{3.85}}{{1.19}} = 3.235cm \\

\Rightarrow DC = BC - BD = 6 - 3.235 = 2.765cm \\

$

Now in triangle ADC

Apply Pythagoras Theorem

$

\Rightarrow {\left( {AC} \right)^2} = {\left( {AD} \right)^2} + {\left( {DC} \right)^2} \\

\Rightarrow {\left( {AC} \right)^2} = {\left( {3.85} \right)^2} + {\left( {2.765} \right)^2} = 22.4677 \\

\Rightarrow AC = \sqrt {22.4677} cm \\

$

So, the area of the triangle is $11.55c{m^2}$and the length of third side is$\sqrt {22.4677} cm$

Note: - In such types of problems always draw the pictorial representation of the given problem, then calculate the perpendicular distance, then calculate the area of triangle using the formula which is stated above, then calculate its third side using Pythagoras Theorem, then we will get the required answer.

Pictorial representation of given problem is shown above

Let, AD is perpendicular to BC.

So, the area of the triangle is half multiply by its perpendicular times base

$ = \dfrac{1}{2}\left( {AD} \right)\left( {BC} \right) = \dfrac{1}{2}\left( {AD} \right)\left( 6 \right) = 3AD$

Now in triangle ABD

$

\sin 50^\circ = \dfrac{{AD}}{{AB}} = \dfrac{{AD}}{5} \\

\Rightarrow AD = 5\sin 50^\circ = 5 \times 0.77 = 3.85cm \\

$

Therefore area of triangle is

Area$ = 3AD = 3 \times 3.85 = 11.55c{m^2}$

Now in triangle ABD

$

\tan 50^\circ = \dfrac{{AD}}{{BD}} = \dfrac{{3.85}}{{BD}} \\

\Rightarrow BD = \dfrac{{3.85}}{{\tan 50^\circ }} = \dfrac{{3.85}}{{1.19}} = 3.235cm \\

\Rightarrow DC = BC - BD = 6 - 3.235 = 2.765cm \\

$

Now in triangle ADC

Apply Pythagoras Theorem

$

\Rightarrow {\left( {AC} \right)^2} = {\left( {AD} \right)^2} + {\left( {DC} \right)^2} \\

\Rightarrow {\left( {AC} \right)^2} = {\left( {3.85} \right)^2} + {\left( {2.765} \right)^2} = 22.4677 \\

\Rightarrow AC = \sqrt {22.4677} cm \\

$

So, the area of the triangle is $11.55c{m^2}$and the length of third side is$\sqrt {22.4677} cm$

Note: - In such types of problems always draw the pictorial representation of the given problem, then calculate the perpendicular distance, then calculate the area of triangle using the formula which is stated above, then calculate its third side using Pythagoras Theorem, then we will get the required answer.