Question

How many triangles does a=6 b=10 A=33° create?

Answer 1

Answer:

2 triangles are possible.

Step-by-step explanation:

Given

a=6

b=10

$\angle$A=33°

To find:

Number of triangles possible ?

Solution:

First of all, let us use the sine rule:

As per Sine Rule:

$\dfrac{a}{sinA}=\dfrac{b}{sinB}$

And let us find the angle B.

$\dfrac{6}{sin33}=\dfrac{10}{sinB}\\sinB = \dfrac{10}{6}\times sin33\\B =sin^{-1}(1.67 \times 0.545)\\B =sin^{-1}(0.9095) =65.44^\circ$

This value is in the 1st quadrant i.e. acute angle.

One more value for B is possible in the 2nd quadrant i.e. obtuse angle which is: 180 - 65.44 = $114.56^\circ$

For the value of $\angle B = 65.44^\circ$, let us find $\angle C$:

$\angle A+\angle B+\angle C = 180^\circ\\\Rightarrow 33+65.44+\angle C = 180\\\Rightarrow \angle C = 180-98.44 = 81.56^\circ$

Let us find side c using sine rule again:

$\dfrac{6}{sin33}=\dfrac{c}{sin81.56^\circ}\\\Rightarrow c = 11.02 \times sin81.56^\circ = 10.89$

So, one possible triangle is:

a = 6, b = 10, c = 10.89

$\angle$A=33°, $\angle$A=65.44°, $\angle$C=81.56°

For the value of $\angle B =$$114.56^\circ$, let us find $\angle C$:

$\angle A+\angle B+\angle C = 180^\circ\\\Rightarrow 33+114.56+\angle C = 180\\\Rightarrow \angle C = 180-147.56 = 32.44^\circ$

Let us find side c using sine rule again:

$\dfrac{6}{sin33}=\dfrac{c}{sin32.44^\circ}\\\Rightarrow c = 11.02 \times sin32.44^\circ = 5.91$

So, second possible triangle is:

a = 6, b = 10, c = 5.91

$\angle$A=33°, $\angle$A=114.56°, $\angle$C=32.44°

So, answer is : 2 triangles are possible.