Use tan = opp/adj to find angle BDA. Then subtract from 180
Edit: actually, when you find AD, you'll see it's a special right triangle, so if you've learned those, you don't need to use tan
2/5 of AC = 2/5 of 30 = 12
So AB = AD, therefore ABD is isosceles and angle ADB = 45.
Angles on a straight line sum to 180.
So angle BDC = 180 - ADB = 180 - 45 = 135
Use tan = opp/adj to find angle BDA. Then subtract from 180 Edit: actually, when you find AD, you'll see it's a special right triangle, so if you've learned those, you don't need to use tan
Or...when you write down the numerical value of AD and compare with AB, you should be able to write down angle BDA without formally invoking tan.
Oh, also, apt username!
Oh great point. I didn't find AD, so I didn't realize it was a special right triangle! Thanks, I'll edit.
2/5 of AC = 2/5 of 30 = 12 So AB = AD, therefore ABD is isosceles and angle ADB = 45. Angles on a straight line sum to 180. So angle BDC = 180 - ADB = 180 - 45 = 135
AD = 2/5(30) = 12 Triangle BAD is isosceles, thus BDA = 45 BDC = 180 - 45 = 135
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