Question

solve the Triangle ABC with a= 1, b=2,c=√3​

Answer 1

Step-by-step explanation:

Let start ,cosA=b²+c²-a²/2bc. Then cos30=(root3)²+1²-a²/2root3*1. Then root3/2=3+1-a²/2root3 on solving we get a=1

Answer 2

This is the correct approach to finding the value of side 'a' in triangle ABC. Using the cosine law, we can find the cosine of angle A by knowing the sides a, b, and c.

The formula for cosine law is: cosA = (b^2 + c^2 - a^2) / (2bc)

Given the values b = 2 and c = √3, we can substitute these values in the formula and find the cosine of angle A.

cosA = (2^2 + (√3)^2 - a^2) / (2 * 2 * √3)

cosA = ((√3)^2 + 1^2 - a^2) / (2 * √3)

cosA = (√3 / 2)

Next, we know that cos30 = √3 / 2. Therefore, we can equate these two values to find the value of 'a'.

cos30 = (√3 / 2) = (2 + 1 - a^2) / (2 * √3)

By solving for a, we get the value of a as 1.

So, the triangle ABC has sides a = 1, b = 2, and c = √3.

For more such questions on triangles: https://brainly.in/question/47901946

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