Question

can someone please tell how to do this question ? I request please no spammers . This is class 11 question . Don't answer if you are not sure !!!​

Answer 1

Step-by-step explanation:

In △ABC Let ∠ACB=θ, BD=x⇒AC=4x

And cosθ=ACBC=4xBC.......(1)

In △BCD,sinθ=BCBD=BCx.....(2) From (1) and (2), we get,

sinθcosθ=41⇒sin2θ=30∘⇒θ=15∘

Therefore, ∠BAC=90∘−15∘=75∘

Hope it will help you bro

Plz Mark me as BRAINLIEST♥️

Answer 2

In △ABC Let ∠ACB=θ, BD=x⇒AC=4x

And cosθ=ACBC=4xBC.......(1)

In △BCD,sinθ=BCBD=BCx.....(2) From (1) and (2), we get,

sinθcosθ=41⇒sin2θ=30∘⇒θ=15∘

Therefore, ∠BAC=90∘−15∘=75

Plz Mark me as BRAINLIEST