- #1

chwala

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- Homework Statement
- See attached ( kindly allow me to post as it is on exam script)

- Relevant Equations
- Understanding of the Triangle

My take:

I got ##BC=10.25## cm, using cosine rule...no issue there. For part (b)

##BK=3cm## using sine rule i.e ##\sin 30^0 =\dfrac{BK}{6}##

Thus it follows that ##∠BDK=48.59^0## ...⇒##∠ADB=131.4^0## correct...any other approach?

Also:

##∠ADB=48.59^0## when BD is on the other side of the given perpendiculor line.

cheers guys

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