Answer:
Q.1.
Sum of angles of a triangle = 180°
∠A + ∠B + ∠c = 180
(3x - 9) + (4x + 15) + 90 = 180
3x - 9 + 4x + 15 + 90 = 180
7x + 96 = 180
7x = 84
x = 12
∠A = 3x - 9 = 3(12) - 9 = 36 - 9 = 27°
∠B = 4x + 15 = 4(12) + 15 = 48 + 15 = 63°
∠A = 27°
∠B = 63°
Q.2.
[tex]using \ pythagoras \ theorem \\x^2 = 7^2 + (7\sqrt{3} )^{2} = 49 + 49(3) = 196\\x = \sqrt{196} = 14[/tex]
[tex]sin y = \frac{opposite}{hypotenuse} = \frac{7\sqrt{3} }{14} = \frac{\sqrt{3} }{2} \\\\y = sin^{-1} (\frac{\sqrt{3} }{2} ) = 60[/tex]
x = 14
y = 60°
