Want to play with angles?
In Δ A B C , let D be the mid point of B C . If ∠ A D B = 4 5 ∘ and ∠ A C D = 3 0 ∘ , determine measure of ∠ B A D in degrees.
The answer is 30.
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2 solutions
mera ho gya :p xD
L e t ∠ B A D = α . S o ∠ A B D = 1 3 5 − α . I n Δ C A D i n t e r n a l ∠ D C A = 4 5 − 1 5 = 3 0 . A p p l y i n g S i n e L a w t o Δ s A D C a n d A D B , S i n 3 0 A D = S i n 1 5 D C = S i n 1 5 B D . . . . . a n d . . . . S i n 1 3 5 − α A D = S i n α B D B u t A D = S i n 1 5 B D S i n 3 0 , . . . S o . . . S i n 1 5 ∗ S i n ( 1 3 5 − α ) B D S i n 3 0 = S i n α B D S i m p l i f y i n g , S i n 1 5 S i n 3 0 = S i n α S i n ( 1 3 5 − α ) = S i n 1 3 5 ∗ C o t α − C o s 1 3 5 . S i n 1 3 5 ∗ C o t α = S i n 1 5 S i n 3 0 + C o s 1 3 5 . S i n c e S i n 2 A = 2 ∗ S i n A ∗ C o s A . ∴ T a n α 1 = 2 ∗ 2 ∗ C o s 1 5 − 1 ∴ α = ∠ B A D = 3 0 o .
Draw BL perpendicular to AC and join L to D . Since angle BCL=30°, we get angle CBL=60°. Since BLC is a right triangle with angle BCL = 30° , we have BL=BC/2=BD. Thus in ∆ BLD , we observe that BL = BD and angle DBL=60°. This implies that BLD is an equilateral ∆ hence LB=LD. Using angle LDB =60° and angle ADB= 45°, we get angle ADL=15°. But angle DAL=15°.Thus LD=LA. We hence have LD=LA=LB. This implies that L is the circumcentre of the ∆. Thus, angle BAD=1/2 angle BLD =1/2*60°=30°.