Question

if triangle ABC , if angle are in A.P. & the b/c= √3/√2 , find all the angle in triangle....​

Answer 1

GIVEN :-

• ∠A, ∠B and ∠C are in A.P.

• B/C = √3/√2

TO FIND :-

The measure of all the angles.

SOLUTION :-

We know, in an A.P.,

B - A = C - B

⇒2B = A + C

Let B = √3x and C = √2x.

⇒2√3x = A + √2x

⇒A = 2√3x - √2x

We know,

A + B + C = 180°         (Angle Sum Property of a Triangle)

⇒2√3x - √2x + √3x + √2x  = 180

⇒3√3x = 180

⇒√3x = 60

⇒x = 60/√3

⇒x = 20√3

$\rule{160}{2}$

Now,

∠B = √3x

⇒∠B = √3 × 20√3

⇒∠B = 20 × 3

⇒∠B = 60°

$\rule{160}{2}$

∠A = 2√3x - √2x

⇒∠A = 2√3 × 20√3 - √2 × 20√3

⇒∠A = 120° - 20√6°

⇒∠A = 120° - 50°  

⇒∠A = 70°

$\rule{160}{2}$

∠C = √2x

⇒∠C = √2 × 20√3

⇒∠C = 20√6

⇒∠C = 50°  

$\rule{160}{2}$

(20√6 is taken approximately as 50°.)

Answer 2

Answer :

A = 50°

B = 60°

C = 70°

Given :

• In triangle ABC , the angles are in A.P.
• B/C = √3/√2

To Find :

• All the angles of ∆ABC

Solution :

According to question all the angles are in A.P. i.e.

$\sf \implies \angle B - \angle A = \angle C - \angle B \\\\ \sf \implies 2\angle B = \angle A + \angle C ....…..(1)$

Again given ,

$\sf \implies \dfrac{B}{C} = \dfrac{\sqrt{3}}{\sqrt{2}}$

$\sf \implies \dfrac{B}{C} = \dfrac{\sqrt{3}}{\sqrt{2}} = x (say)$

$\sf \implies \dfrac{B}{\sqrt{3} }= \dfrac{C}{\sqrt{2}} = x \\\\ \sf \implies B = \sqrt{3}x , \: \: C = \sqrt{2}x$

Putting the values in (1) we have :

$\sf \implies 2\sqrt{3}x = A + \sqrt{2}x \\\\ \sf \implies A = 2\sqrt{3}x - \sqrt{2}x$

Now from the properties of triangles we have:

Sum of all angels of a triangle = 180°

$\sf \implies \angle A + \angle B + \angle C = 180\degree \\\\ \sf \implies 2\sqrt{3}x - \sqrt{2}x + \sqrt{3}x + \sqrt{2}x = 180\degree \\\\ \sf \implies 3\sqrt{3} = 180\degree \\\\ \sf \implies x = \dfrac{180\degree}{3\sqrt{3}} \\\\ \sf \implies x = 20\sqrt{3}$

Thus , the angles are :

$\sf \bullet \: \: \angle A = 2\sqrt{3}\times 20\sqrt{3} - \sqrt{2}\times 20\sqrt{3} \\\\ \sf \implies \angle A = 70\degree \: \: (approx) \\\\ \sf \bullet \: \: \angle C = \sqrt{2}\times 20\sqrt{3} \\\\ \sf \implies \angle C = 50\degree \: \: (approx) \\\\ \sf\bullet \: \: \angle B = 20\sqrt{3}\times \sqrt{3} \\\\ \sf \implies \angle B = 60\degree$