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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