Question

Find the perimeter of each of the two non congruent triangles where a=15,b=20 and a=29

Answer 1

Answer:

b. about 63.9 units and 41.0 units

Step-by-step explanation:

In question ∠a= 29° and Side of a= 15 and b= 20

Using sine rule of congruence of triangle.

⇒ $\frac{a}{sin A} = \frac{b}{sin B} = \frac{c}{sin C}$

⇒ $\frac{15}{Sin 29} = \frac{20}{sin B}$

Using value of sin 29°

⇒ $\frac{15}{0.49} = \frac{20}{sin B}$

Cross multiplying both side.

⇒ Sin B= $\frac{20\times 0.49}{15} = 0.65$

∴ B= 41°

Now, we have the degree for ∠B= 41°.

Next, lets find the ∠C

∵ we know the sum total of angle of triangle is 180°

∴∠A+∠B+∠C= 180°

⇒ $29+41+B= 180$

subtracting both side by 70°

∴∠C= 110°

Now, again using the sine rule to find the side of c.

$\frac{b}{SinB} = \frac{c}{SinC}$

⇒$\frac{20}{sin41} = \frac{C}{sin110}$

Using the value of sine and cross multiplying both side.

⇒ C= $\frac{20\times 0.94}{0.65} = 28.92$

∴ Side C= 28.92.

Now, finding perimeter of angle of triangle

Perimeter of triangle= a+b+c

Perimeter of triangle= $(15+20+28.92)= 63.9$

∴ Perimeter of triangle= 63.9 units