C-09-Q081mediumsingle_mcqIn right △ABC\triangle ABC△ABC with ∠C=90∘\angle C = 90^\circ∠C=90∘, AB=13AB = 13AB=13, BC=12BC = 12BC=12, ∠ABC=θ\angle ABC = \theta∠ABC=θ. Then AC=?AC = ?AC=?a555b777c111111d121212ব্যাখ্যা∠C=90∘\angle C = 90^\circ∠C=90∘ makes ABABAB the hypotenuse, so by Pythagoras AC=AB2−BC2=132−122=169−144=25=5AC = \sqrt{AB^2 - BC^2} = \sqrt{13^2 - 12^2} = \sqrt{169 - 144} = \sqrt{25} = 5AC=AB2−BC2=132−122=169−144=25=5.