Question

One angle of an isosceles triangle is 150 degrees. Ifthe

Answer 1

Answer:

Perimeter would be 23.57 ft ( approx)

Step-by-step explanation:

Consider triangle ABC is an isosceles triangle,

In which AB = AC,

And, m∠A = 150°,

∵ AB = AC ⇒ m∠B = m∠C,

Now sum of all interior angles of a triangle is 180°,

i.e. m∠A + m∠B + m∠C = 180°,

150°+ m∠B + m∠B= 180°,

2m∠B + 150° = 180°

2m∠B = 30°

⇒ m∠B = 15°

Let D ∈ BC such that AD ⊥ BC,

∵ Altitude of an isosceles triangle is its median,

i.e, BD = DC or BD = $\frac{1}{2}$ BC

In triangle ADB,

tan 15° = $\frac{AD}{BD}$

$\implies \tan 15^{\circ}=\frac{AD}{\frac{1}{2}BC}=\frac{2AD}{BC}$

$\implies AD = \frac{BC \tan 15^{\circ}}{2}$              .............(1)

Now, area of triangle ABC = $\frac{1}{2}\times AD\times BC$

If area = 9 square ft,

$\frac{1}{2}\times AD\times BC = 9$

From equation (1),

$\frac{1}{2}\times \frac{BC \tan 15^{\circ}}{2}\times BC = 9$

$BC^2\tan 15^{\circ} = 36$

$BC^2 =\frac{36}{\tan 15^{\circ}}=134.35$

$\implies BC = 11.59\text{ ft}$

From equation (1),

$AD = \frac{11.59 \tan 15^{\circ}}{2}=1.55$

Using Pythagoras theorem,

$AB = \sqrt{AD^2 + BD^2}=\sqrt{1.55^2+(\frac{11.59}{2})^2}=5.99\text{ ft}$

Hence, perimeter of the triangle ABC= AB + BC + CA

= 5.99 + 11.59 + 5.99

= 23.57 ft