Question

One side of a triangle is 6 centimetres and the angles at its ends are 40°
and 65º. Calculate its area.​

Answer 1

Step-by-step explanation:

Let ABC be the triangle in which BC = 6 cm. <B=40 deg and <C=65 deg.

So <A = 180–40–65 = 75 deg.

a/sin A = b/sin B = c/sin C, or

6/sin 75 = b/sin 40 = c/sin 65, or

b = 6*sin 40/sin 75 = 3.992776208 cm

c = 6*sin 65/sin 75 = 5.629673184 cm.

Area of ABC can be found by four methods.

Method 1: Area = (AB*AC*sin A)/2 = [5.629673184*3.992776208*sin 75]/2 = 10.85605251 sq cm.

Method 2: Area = (BC*BA*sin B)/2 = [6*5.629673184*sin 40]/2 = 10.85605251 sq cm.

Method 3: Area = (CA*CB*sin C)/2 = [3.992776208*6*sin 65]/2 = 10.85605251 sq cm.

Methods 1, 2 and 3 are basically the same that is area = base*height/2, except the bases are different in the three methods.

Method 4: 2s = a+b+c =6+3.992776208+5.629673184 = 15.62244939, and s = 7.811224696.

Area by Heron’s formula =[7.811224696(7.811224696- 6)(7.811224696-3.992776208)(7.811224696-5.629673184)]^0.5

= [7.811224696*1.811224696*3.818448488*2.101551512]^0.5

= 117.8538761^0.5

= 10.85605251 sq cm.

Please mark as brainliest answer.............

Answer 2

Given,

The length of one side of a triangle$\[ = 6{\rm{cm}}\]$

The first angle of the triangle$\[ = 40^\circ \]$

The second angle of the triangle$\[ = 65^\circ \]$

Solution,

Refer to figure.

In triangle APB,

$\[\begin{array}{l} \Rightarrow \tan 40^\circ = \frac{h}{x}\\ \Rightarrow x = \frac{h}{{\tan 40^\circ }}\\ \Rightarrow x = \frac{h}{{0.8391}} - - - - - \left( 1 \right)\end{array}\]$

In triangle APC,

$\[\begin{array}{l} \Rightarrow \tan 65^\circ = \frac{h}{{6 - x}}\\ \Rightarrow 2.1445 = \frac{h}{{6 - \frac{h}{{0.8391}}}}\\ \Rightarrow 2.1445 \times \left( {6 - \frac{h}{{0.8391}}} \right) = h\\ \Rightarrow 12.867 - 2.556h = h\end{array}\]$

Solve further,

$\[\begin{array}{l} \Rightarrow 3.556h = 12.867\\ \Rightarrow h = 3.618cm\end{array}\]$

Therefore, the area of the triangle is

$\[\begin{array}{l} \Rightarrow \frac{1}{2} \times 6 \times 3.618\\ \Rightarrow 3 \times 3.618\\ \Rightarrow 10.854{\rm{c}}{{\rm{m}}^2}\end{array}\]$

Hence, the area of the triangle is $\[10.854{\rm{c}}{{\rm{m}}^2}\]$