Question: One angle of a triangle is 72° and rest two is in the ratio 1:3. Find all the angles in grade.

Answer: $80^g, 30^g, 90^g$

Solution:
In a triangle,
one angle ($\measuredangle A$) = 72° 
Let the other angles be $\angle B and \angle C$ respectively.


Given,

$\measuredangle B : \measuredangle C = 1:3$

$or, \dfrac{ \measuredangle  B}{ \measuredangle  C} = \dfrac{1}{3}$

$or, 3 \measuredangle B = \measuredangle C$

$or, \measuredangle C = 3 \measuredangle B$


And,

$ \measuredangle A^g = \dfrac{ \measuredangle A° ×200}{180}$

$or, \measuredangle A^g = \dfrac{ 72° × 200}{180}$

$\therefore \measuredangle A^g = 80^g$


We know,

Sum of interior angles of a triangle is $200^g$.

$\measuredangle A^g + \measuredangle B^g + \measuredangle C^g = 200^g$

$or, 80^g + \measuredangle B^g + 3\measuredangle B^g = 200$

$or, 4 \measuredangle B^g = 120^g$

$or, \measuredangle B^g = \dfrac{120^g}{4}$

$\therefore \measuredangle B^g = 30^g$


Again,

$\measuredangle C = 3 \measuredangle B$

$or, \measuredangle C^g = 3 \measuredangle B^g$

$or, \measuredangle C^g = 3×30^g$

$\therefore \measuredangle C^g = 90^g$



Hence, the required angles of the triangle in grades are $80^g, 30^g, 90^g$

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