A geometry problem by Dinesh Nath Goswami

Geometry Level 2

In a triangle ABC, a=15, b=36, c=39 then sin ⁡ C 2 \sin { \frac { C }{ 2 } } =?

− 3 2 -\frac { \sqrt { 3 } }{ 2 } 1 2 \frac { 1 }{ 2 } 1 2 \frac { 1 }{ \sqrt { 2 } } 3 2 \frac { \sqrt { 3 } }{ 2 }

Dinesh Nath Goswami by Dinesh Nath Goswami , 20, India

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2 solutions

a, b and c form a Pythagorean triplet, so they are the sides of a right triangle.

Here c is the largest side, so angle C is the greatest angle (=90)

Now sin ⁡ C 2 \sin { \frac { C }{ 2 } } = sin ⁡ 45 \sin { 45 } = 1 2 \frac { 1 }{ \sqrt { 2 } }

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Assuming that we don't know that it is a pythagorean triplet, we use the law of cosines to find for ∠ C \angle C .

3 9 2 = 1 5 2 + 3 6 2 − 2 ( 15 ) ( 36 ) ( cos ⁡ C ) 39^2=15^2+36^2-2(15)(36)(\cos C)

C = 9 0 ∘ C=90^\circ

C 2 = 90 2 = 45 \dfrac{C}{2}=\dfrac{90}{2}=45

sin ⁡ 45 = 1 2 \sin 45 = \dfrac{1}{\sqrt{2}}

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