Question

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

Answer 1

2 triangles are possible.

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:

`(a)/(sinA)=(b)/(sinB)`

And let us find the angle B.

`(6)/(sin33)=(10)/(sinB)\\sinB = (10)/(6)* sin33\\B =sin^(-1)(1.67 * 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:

`(6)/(sin33)=(c)/(sin81.56^\circ)\\\Rightarrow c = 11.02 * 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:

`(6)/(sin33)=(c)/(sin32.44^\circ)\\\Rightarrow c = 11.02 * 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.