Answer:
:
(a) Solution: 1 Mark
Type of triangle: 1 Mark
(b) Solution: 1 Mark
Correct answer: 1 Mark
(a) For a triangle, sum of all angles = 180${}^o$
Or, (2x - 15${}^o$) + (x + 20${}^o$) + (x + 15${}^o$) = 180${}^o$
Or, 2x + x + x - 15${}^o$ + 20${}^o$ + 15${}^o$ = 180${}^o$
Or, 4x + 20${}^o$ = 180${}^o$
Or, 4x = 160${}^o$
Or, x = $\frac{160}{4}$ = 40${}^o$
The angles of the triangle will be, (2x - 15${}^o$ = 65${}^o$), (x + 20${}^o$ = 60${}^o$) and (x + 15${}^o$ = 55${}^o$)
Since all angles are less than 90${}^o$, it's an 'acute angled triangle' and also all the angles are of different values therefore all the sides will be of different length, hence it is a 'scalene triangle'.
(b

In triangle ABD
$\Rightarrow 2x+4+90+30=180$
$\Rightarrow 2x=180-124$
$\Rightarrow x=\frac{56}{2}$
$\Rightarrow x=28$
In triangle ADC
$\Rightarrow 4x-k+90+10=180$
$\Rightarrow 4x-k=180-100$
$\Rightarrow 4\left(28\right)-k=80$ (Substituting the value of $x=28$)
$\Rightarrow k=4\left(28\right)-80$
$\Rightarrow k=112-80$
$\Rightarrow k=32$
$\therefore$ The angles are ${40}^{\circ },{60}^{\circ }and{80}^{\circ }$ for the given triangle.