Question

if in a ∆ABC, angle A =30° and angle B = 45° ,then find the value of cot C.​

Answer 1

Here is a solution without any trigonometry or calculations.

Let AH be the altitude of triangle ΔABC through vertex A , i.e. the segment AH is perpendicular to AC with H on AC . The angle ∠BCH=∠BCA=30∘ in the right-angled triangle BCH yields that ∠CBH=60∘ . Furthermore, D is the midpoint of segment BC , so HD is the median of the right angled triangle ΔBCH which allows us to conclude that triangle BDH is equilateral and that BD=CD=HD=HB . Moreover, ∠CHD=30 .

Angle chasing in triangle ΔACD shows that ∠HAD=∠CAD=15∘ . But together with that,

∠HDA=∠HDB−∠ADB=60∘−45∘=15∘

Since ∠HDA=15∘=∠HAD , triangle ΔADH is isosceles with HA=HD . Therefore

HA=HB=HD

Construct the circle centered at point H and radius HA=HB=HD . Then this circle passes through the points A,B and D . From the relation between central and peripheral angles in a circle

∠BAD=12∠BHD=1260∘=30∘

Answer 2

$\underline\red{\bold{answer}}$

$\cot(c) = - 2 + \sqrt{3}$

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${\bold{\boxed{\sf{\green{GIVEN \: THAT }}}}}</p><p>$

• In a triangle ∆ABC , angle A = 30° and angle B = 45°

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$\underline\red{\bold{FIND}}$

• What is the value of Cot C ?

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$\underline\red{\bold{SOLUTION}}$

• In a Triangle the sum of three angle = 180°

A + B + C = 180°

30° + 45° + C = 180°

C = 180° - 75°

C = 105°

Now the value of

$\cot(c) = \cot(105°) = - 2 + \sqrt{3}$