MCQ
If $\triangle\text{ABC}$ is right angled at $C,$ then the value of $\cos(\text{A}+\text{B})$ is :
- ✓$0$
- B$1$
- C$\frac{1}{2}$
- D$\frac{\sqrt{3}}{2}$
✓
Answer
Correct option: A.
$0$
In a right angled traingle $\text{ABC}, \triangle\text{C}$ is a righta angle.
We know that, the sum of angles of a triangle is $180^\circ .$
$\therefore\ \angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
$\Rightarrow\ \angle\text{A}+\angle\text{B}+90^\circ=180^\circ$
$\Rightarrow\ \angle\text{A}+\angle\text{B}=90^\circ$
$\therefore\ \cos(\text{A}+\text{B})=\cos90^\circ=0$
Hence, the correct answer is option $(a).$
We know that, the sum of angles of a triangle is $180^\circ .$
$\therefore\ \angle\text{A}+\angle\text{B}+\angle\text{C}=180^\circ$
$\Rightarrow\ \angle\text{A}+\angle\text{B}+90^\circ=180^\circ$
$\Rightarrow\ \angle\text{A}+\angle\text{B}=90^\circ$
$\therefore\ \cos(\text{A}+\text{B})=\cos90^\circ=0$
Hence, the correct answer is option $(a).$
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
The common difference of the $A.P$ whose an $= -3n + 7$ is :
→A vertical stick $1.8m$ long casts a shadow $45\ cm$ long on the ground. At the same time, what is the lenght of the shadow of a pole $6m$ high?
→The point on the $y-$ axis which is equidistant from the points $(6, 5)$ and $(-4, 3)$ is :
→The mean of $20$ numbers is zero. of them, at the most, how many may be greater than zero?
→If the radii of the circular ends of a bucket of height $40\ cm$ are of lengths $35\ cm$ and $14\ cm,$ then the volume of the bucket in cubic centimeters, is:
→In the given figure, AB and AC are tangent to the circle with centre O such that $\angle\text{BAC}=40^\circ.$ Then, $\angle\text{BOC}$ is equal to:
→If $\theta=45^{\circ}$ then the value of $\frac{1-\cos 2 \theta}{\sin 2 \theta}$ is -
→The sum of first 16 terms of the A.P.: $10,6,2, \ldots$, is
→The hour hand of a clock is $6\ cm$ long. The area swept by it between $11.20 \text{am}$ and $11.55 \text{am}$ is :
→Mark the correct alternative in the following : The first three terms of an $A.P$. respectively are $3y - 1, 3y + 5$ and $5y + 1$. Then $, y$ equals :
→