Question

In the adjacent figure, angleCAB || angleBAD = 1 : 2 Find the alternet internal angles of triangle ABC.​

Answer 1

*Answer:-

Internal angles of ΔABC is 40°.

Given:-

To find all the internal angles of triangle ABC.

From the figure,

→∠ABC = x

→∠BCA = x+3

→∠CAB : ∠ BAD = 1 : 2

Angle ∠CAB and ∠BAD are in the ratio of 1 : 2.

Sum of all angles is 180°.

→∠ABC + ∠BCA + ∠CAB = 180°

→x + x + 13° + ∠CAB = 180°

→2x + 13° + ∠CAB = 180°

→2x + ∠CAB = 180° - 13°

→∠CAB = 167° - 2x

Also in supplementary,

→∠CAB + ∠DAE + ∠BAD = 180°

→( 167° - 2x ) + 60° + ∠BAD = 180°

→167° -2x + ∠BAD = 180° - 60°

→∠BAD = 120° - 167° + 2x = 2x - 47°

→∠BAD = 2x - 47°

Where, ∠CAB / ∠BAD = 1 / 2

Substituting, ∠CAB and ∠BAD values, it gives:-

$= > \: \frac{167 - 2x}{2x - 47} = \frac{1}{2}$

→334° - 4x = 2x - 47°

→334° + 47° = 2x + 4x

→6x = 381°

→x = (127 / 2)°

Angles of ΔABC,

∠ABC = x = (127 / 2)°

∠BCA = x+13°

= (127 / 2)° + 13°

= (153 / 2)°

∠CAB = 167° - 2x

= 167° - 2(127 / 2)°

= 40°

Therefore, internal angles of ΔABC is 40°.