Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

Length of two sides of a triangle are $6cm$and$5cm$. If their included angle is$50^\circ $. What is the area of the triangle? Also find the length of its third side.
$\left( {\sin 50^\circ = 0.77,\cos 50^\circ = 0.64,\tan 50^\circ = 1.19} \right)$

seo-qna
Last updated date: 27th Mar 2024
Total views: 420.6k
Views today: 7.20k
MVSAT 2024
Answer
VerifiedVerified
420.6k+ views
Hint:Area of the triangle is half multiplied by its perpendicular times base.
seo images


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.
Recently Updated Pages