Question

In triangle ABC, right angled at B, if one angle is ${45}^{\circ }$, find the value of sin A and cos C, without using the table.

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

Answer 1

The correct option is C

$\frac{1}{\sqrt{2}}$, $\frac{1}{\sqrt{2}}$

In right angle triangle ABC

Let $\angle A$ = ${45}^{\circ }$, $\angle B$ = ${90}^{\circ }$

Then $\angle C$ = ${180}^{\circ }-{90}^{\circ }-{45}^{\circ }$ = ${45}^{\circ }$

Let AB = BC = a [since $\angle B$ and $\angle C$ = ${45}^{\circ }$, its an isosceles right triangle]

Apply Pythagoras theorem in $△$ABC

AC = $\sqrt{a^2+a^2}$ = a$\sqrt{2}$

Using Trigonometric ratio

sin A = sin ${45}^{\circ }$ = $\frac{BC}{AC}$ = $\frac{a}{a\sqrt{2}}$ = $\frac{1}{\sqrt{2}}$

cos C = cos ${45}^{\circ }$ = $\frac{BC}{AC}$ = $\frac{a}{a\sqrt{2}}$ = $\frac{1}{\sqrt{2}}$