Question

calculate the perimeter and area of the triangle in the picture ​

Answer 1

As the sum of all angles in triangle is 180 hence,60 +75+x=180 and the side is 45

Answer 2

$\large\underline{\bold{Solution :- }}$

Given :-

Let us assume a triangle ABC in which

• AB = 'c' cm

• AC = 'b' cm

• BC = b = 4 cm

• ∠B = 60°

• ∠C = 75°

To Find :-

• Area of triangle ABC.

Concept used :-

Angle sum Property of triangle

• The sum of angles of a triangle is 180°.

Sine Law

Let us consider a triangle ABC whose sides AB = c units, BC = a units and AC = b units, then

$\rm :\longmapsto\:\dfrac{a}{sinA} = \dfrac{b}{sinB} = \dfrac{c}{sinC}$

Area of triangle ABC is given by

$\rm :\longmapsto\:Area_{(triangle)} = \dfrac{1}{2} ab \: sinC$

CALCULATION :-

We know,

• In triangle ABC,

• Sum of the angmles is 180°.

So,

• ∠A + ∠B + ∠C = 180°.

• ∠A + 60° + 75° = 180°

• ∠A = 180° - 135°

• ∠A = 45°

Now,

• Using Sine Law, In triangle ABC , we have

$\rm :\longmapsto\:\dfrac{a}{sinA} = \dfrac{b}{sinB}$

$\rm :\longmapsto\:\dfrac{4}{sin45 \degree \: } = \dfrac{b}{sin60\degree \:}$

$\rm :\implies\:b \: = \dfrac{4}{sin45\degree \:} \times sin60\degree \:$

$\rm :\implies\:b \: = 4 \times \sqrt{2} \times \dfrac{ \sqrt{3} }{2}$

$\rm :\implies\:b \: = \: 2 \sqrt{6} \: cm$

Now,

we know,

$\rm :\longmapsto\:sin(x + y) = sinx \: cosy \: + \: siny \: cosx$

So,

$\rm :\longmapsto\:sin \: 75\degree \: = sin(45\degree \: + \: 30\degree \:)$

$\rm :\longmapsto\:sin75\degree \: = sin45\degree \:cos30\degree \: + sin30\degree \:cos45\degree \:$

$\rm :\longmapsto\:sin75\degree \: = \dfrac{1}{ \sqrt{2} } \times \dfrac{ \sqrt{3} }{2} + \dfrac{1}{2} \times \dfrac{1}{ \sqrt{2} }$

$\rm :\longmapsto\:sin75\degree \: = \dfrac{ \sqrt{3} + 1}{2 \sqrt{2} }$

Now,

• Area of triangle ABC is evaluated as

$\rm :\longmapsto\:Area_{(triangle)} = \dfrac{1}{2} ab \: sinC$

$\rm :\longmapsto\:Area_{(triangle)} = \dfrac{1}{2} \times 4 \times 2 \sqrt{6} \times sin75\degree \:$

$\rm :\longmapsto\:Area_{(triangle)} = 4 \sqrt{6} \times \bigg(\dfrac{ \sqrt{3} + 1 }{2 \sqrt{2} } \bigg)$

$\rm :\longmapsto\:Area_{(triangle)} = 2 \sqrt{3} \times ( \sqrt{3} + 1)$

$\rm :\longmapsto\:Area_{(triangle)} = 6 + 2 \sqrt{3} \: {cm}^{2}$

Additional Information :-

Cosine Law

$\boxed{ \tt \:cosA = \dfrac{ {b}^{2} + {c}^{2} - {a}^{2} }{2bc} }$

$\boxed{ \tt \:cosB = \dfrac{ {c}^{2} + {a}^{2} - {b}^{2} }{2ac} }$

$\boxed{ \tt \:cosC = \dfrac{ {a}^{2} + {b}^{2} - {c}^{2} }{2ab} }$

Projection Formula :-

$\boxed{ \tt \:b \: = a \: cosC \: + \: c \: cosA }$

$\boxed{ \tt \: a = b \: cosC \: + \: c \: cosB}$

$\boxed{ \tt \: c = a \: cosB \: + \: b \: cosA}$

Law of Tangents :-

$\boxed{ \tt \: tan \bigg( \dfrac{B - C}{2} \bigg) = \dfrac{b - c}{b + c} cot\dfrac{A}{2} }$

$\boxed{ \tt \: tan \bigg( \dfrac{C - A}{2} \bigg) = \dfrac{c - a}{c + a} cot\dfrac{B}{2} }$

$\boxed{ \tt \: tan \bigg( \dfrac{A - B}{2} \bigg) = \dfrac{a - b}{a + b} cot\dfrac{C}{2} }$