Question

sinA=sin^2B & 2cos^2A=3cos^2B then the triangle ABC is​

Answer 1

Answer:

Step-by-step explanation:

sinA=sin2B→(1)  

2cos2A=3cos2B

⇒2cos2A=3(1−sin2B)

From (1),

⇒2(1−sin2A)=3(1−sinA)

⇒2sin2A−3sinA+1=0

⇒2sin2A−2sinA−sinA+1=0

⇒2sinA(sinA−1)−1(sinA−1)=0

⇒(2sinA−1)(sinA−1)=0

⇒sinA=1orsinA=12

⇒A=90∘orA=30∘

When A=90∘, sin2B=1⇒B=90∘, which is not possible for a triangle.

When A=30∘, sin2B=12⇒B=45∘

⇒C=(180−30−45)∘=75∘

∴ΔABC is an obtuse angle triangle.

Answer 2

Explanation:

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