Question

What is the area of the triangle ABC with sides a =10 cm, c =4 cm and angle B =30 degree?
a) 16cm^2
6) 12cm²
c) l0cm²
d) 8 cm^2​

Answer 1

Answer:

Step-by-step explanation:

there is a formula which is

area of triangle = 1/2absin theta

so we put all values in this formula

  = 1/2 *10*4*sin 30

 = 1/2*40*1/2

  =10 cm

Answer 2

$\large\underline{\sf{Solution-}}$

Given that,

In triangle ABC, a = 10 cm, c = 4 cm and ∠B = 30°

Let us drop a perpendicular from vertex A on line BC intersecting BC at D.

Now, In triangle ABD

$\rm :\longmapsto\:sin30 \degree \: = \: \dfrac{AD}{AB}$

$\rm :\longmapsto\:\dfrac{1}{2} \: = \: \dfrac{AD}{4}$

$\rm :\longmapsto\:\dfrac{4}{2} \: = \:AD$

$\bf\implies \:AD = 2 \: cm$

Now,

$\rm :\longmapsto\:Area_{\triangle ABC} = \dfrac{1}{2} \times AD \times a$

$\rm :\longmapsto\:Area_{\triangle ABC} = \dfrac{1}{2} \times 2 \times 10$

$\bf :\longmapsto\:Area_{\triangle ABC} = 10 \: {cm}^{2}$

Hence,

• Option (c) is correct

Remark :-

$\red{ \boxed{ \sf{ \:Area_{\triangle ABC} = \frac{1}{2}absinC}}}$

$\red{ \boxed{ \sf{ \:Area_{\triangle ABC} = \frac{1}{2}bcsinA}}}$

$\red{ \boxed{ \sf{ \:Area_{\triangle ABC} = \frac{1}{2}casinB}}}$

Additional Information :-

$\red{ \boxed{ \sf{ \frac{a}{sinA} = \frac{b}{sinB} = \frac{c}{sinC} = k }}}$

$\red{ \boxed{ \sf{ \:cosA = \frac{ {b}^{2} + {c}^{2} - {a}^{2} }{2bc}}}}$

$\red{ \boxed{ \sf{ \:cosB = \frac{ {c}^{2} + {a}^{2} - {b}^{2} }{2ca}}}}$

$\red{ \boxed{ \sf{ \:cosC = \frac{ {a}^{2} + {b}^{2} - {c}^{2} }{2ab}}}}$