Question

please solve it ..find the value of 4m and 3m​

Answer 1

Answer:

The base angles can be written as - [180° - 4m] and [180° - 3m] --- [Linear Pairs]

100° + [180° - 4m] + [180° - 3m] = 180°  ---[Angle sum property of triangles]

460° - 7m = 180°

-7m = -280

7m = 280

m = $\frac{280}{7}$

m = 40°

Therefore, the base angles are 60° and 20°.

3m = 120°

4m = 160°

Hope it helps!

Answer 2

GIVEN :-

• Angle A = 100°

TO FIND :

• The value of 4m and 3m

SOLUTION :-

$\setlength{ \unitlength}{20} \begin{picture}(0,0) \put(1,1){\vector(-1,0){5}}\put(1,1){\vector(1,0){5}}\put( - 2,1){\line(1,1){2.8}}\put( - 1.5,1){\oval(1.5,1)[ tl ]}\put( 3.5,1){\line( - 1,1){2.8}}\qbezier(4,1)(4,1.5)(3,1.5)\qbezier(0,3.)(1,2.55)(1.5,3)\put(0.2,2.3){ \displaystyle \tt {100}^{\circ} }\put( - 2.5, 1.7){ \tt 4m }\put(3.8, 1.7){ \tt 3m }\qbezier( - 1.5,1.5)( 0,0.8)( - 1.9,1)\put(0.5,4){ \tt A }\put( - 2,0.5){ \tt B }\put(3.5,0.5){ \tt C }\end{picture}$In In ABC,$\implies \bf \: 4m \: + \: \angle \: B = 180 \degree \: \: \: \: \: \: [ \because \: linear \: pair] \\ \\ \implies \bf \angle \: B = \: 180 \degree - 4m \: .......(1)$Now,$\implies \bf \: 3m = 100 \degree + \angle \: B \: \: \: \: \: \: [ \because \: exterior \: \angle \: \: property] \\ \\ \implies \bf \: 3m \: = 100\degree + 180\degree - 4m \\ \\ \implies \bf \: 3m + 4m = 280\degree \\ \\ \implies \bf \: 7m = 280\degree \\ \\ \implies \bf \: m \: = \frac{280\degree}{7} \\ \\ \implies \bf \: m = 40 \degree$

Hence we found m = 40°

Therefore,

▪︎4m = 4 × 40° = 160°

▪︎3m = 3 × 40° = 120°