C-09-Q071mediumsingle_mcq

In right $\triangle ABC$ with $\angle C = 90^\circ$ , $AB = 29$ , $BC = 21$ , $\angle ABC = \theta$ . Then $AC = ?$

a $18$
b $20$
c $22$
d $24$

ব্যাখ্যা

With $\angle C = 90^\circ$ , $AB$ is the hypotenuse. By Pythagoras, $AC = \sqrt{AB^2 - BC^2} = \sqrt{29^2 - 21^2} = \sqrt{841 - 441} = \sqrt{400} = 20$ .

এই অধ্যায়ের আরও প্রশ্ন

C-09-Q072With the same data (, , , ),
C-09-Q073With the same data,
C-09-Q074With the same data,
C-09-Q075If ,
C-09-Q076If , then
C-09-Q077If , then opposite adjacent
C-09-Q078If , hypotenuse (taking opposite , adjacent )
C-09-Q079If ,
C-09-Q080If ,
C-09-Q081In right with , , , . Then

Trigonometric Ratio — সব প্রশ্ন দেখুন (217টি) →

In right $\triangle ABC$ with $\angle C = 90^\circ$ , $AB = 29$ , $BC = 21$ , $\angle ABC = \theta$ . Then $AC = ?$

18

20

22

24

ব্যাখ্যা

With $\angle C = 90^\circ$ , $AB$ is the hypotenuse. By Pythagoras, $AC = \sqrt{AB^2 - BC^2} = \sqrt{29^2 - 21^2} = \sqrt{841 - 441} = \sqrt{400} = 20$ .

In right △ABC\triangle ABC△ABC with ∠C=90∘\angle C = 90^\circ∠C=90∘, AB=29AB = 29AB=29, BC=21BC = 21BC=21, ∠ABC=θ\angle ABC = \theta∠ABC=θ. Then AC=?AC = ?AC=?

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

In right △ABC\triangle ABC△ABC with ∠C=90∘\angle C = 90^\circ∠C=90∘, AB=29AB = 29AB=29, BC=21BC = 21BC=21, ∠ABC=θ\angle ABC = \theta∠ABC=θ. Then AC=?AC = ?AC=?

ব্যাখ্যা

এই অধ্যায়ের আরও প্রশ্ন

In right $\triangle ABC$ with $\angle C = 90^\circ$ , $AB = 29$ , $BC = 21$ , $\angle ABC = \theta$ . Then $AC = ?$

In right $\triangle ABC$ with $\angle C = 90^\circ$ , $AB = 29$ , $BC = 21$ , $\angle ABC = \theta$ . Then $AC = ?$