Question

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

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

Answer 1

The correct option is B $\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}},1,1$

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
$sinA=sin{45}^{\circ }=BC/AC=\frac{a}{a\sqrt{2}}=\frac{1}{sqrt2}cosC=sin{45}^{\circ }=AB/AC=\frac{a}{a\sqrt{2}}=\frac{1}{sqrt2}cotA=cot{45}^{\circ }=AB/BC=\frac{a\sqrt{2}}{a\sqrt{2}}=1tanC=tan{45}^{\circ }=AB/BC=\frac{a\sqrt{2}}{a\sqrt{2}}=1$