Question

In triangle ABC, right angled at B, if one angle is ${45}^{\circ }$, find the value of sin A, cosC, cot A and tan C respectively.

• A. $1,1,1/\sqrt{3},1/\sqrt{3}$
• B. $2,2,\sqrt{2},\sqrt{2}$
• C. $1/\sqrt{2},1/\sqrt{2},1,1$
• D. $1/\sqrt{3},1/\sqrt{3},1,1$

Answer 1

The correct option is C $1/\sqrt{2},1/\sqrt{2},1,1$

In right angle triangle ABC, Sin A = BC/AC, Sin C = AB/AC


In right angle triangle ABC
Let $\angle A={45}^{\circ },$ angle $\angle B={90}^{\circ }$
Then $\angle C={180}^{\circ }-{90}^{\circ }-{45}^{\circ }={45}^{\circ }$
Let AB = BC = a
Apply Pythagoras theorem in $\Delta ABC$
$AC=\sqrt{(a^2+a^2)}=a\sqrt{2}$
Using Trigonometric ratio
sin A = sin ${45}^{\circ }=BC/AC=a/a\sqrt{2}=1/\sqrt{2}cosC=cos{45}^{\circ }=BC/AC=a/a\sqrt{2}=1/\sqrt{2}cotA=cot{45}^{\circ }=AB/BC=a\sqrt{2}/a\sqrt{2}=1tanC=tan{45}^{\circ }=AB/BC=a\sqrt{2}/a\sqrt{2}=1$