Question
In a rhombus ABCD, if $\angle\text{ACB}=40^\circ,$ then $\angle\text{ADB}=$
- 70°
- 45°
- 50°
- 60°
✓
Answer
- 50°
Solution:
Consider $\triangle\text{AOD} \ \&\ \triangle\text{COB}$ $$
$\angle\text{AOD}=\angle\text{COB}=90^\circ$
AD = BC (Sides of Rhombus)
AO = CO (Diagonals bisects each other)
So by RHS property, $\triangle\text{AOD}\cong\triangle\text{COB}$
$\Rightarrow\angle\text{OAD}=\angle\text{OCB}=40^\circ$
$\angle\text{ADB}=\angle\text{ADO}=180^\circ-90^\circ-40^\circ=50^\circ$
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
In a frequency distribution, ogives are graphical representation of:
If I is the length of a diagonal of a cube of volume V, then.
The points $O(0,0), A(4,0)$ and $B(0,4)$
→→
The graph of x = 4 is a line.
Two parallelograms are on the same base and between the same parallels. The ratio of their areas is:
- 1 : 2
- 2 : 1
- 1 : 1
- 3 : 1
In the given figure, AD is a median of $\triangle\text{ABC}$ and E is the mid-point of AD. If BE is joined and produced to meet AC in F, then AF = ?
→- $\frac{1}{2}\text{AC}$
- $\frac{1}{3}\text{AC}$
- $\frac{2}{3}\text{AC}$
- $\frac{3}{4}\text{AC}$
In the following, write the correct answer.
In $\triangle\text{PQR},$ if $\angle\text{R}>\angle\text{Q}$ then:
→In $\triangle\text{PQR},$ if $\angle\text{R}>\angle\text{Q}$ then:
In $\triangle\text{ABC,}$ if $\angle\text{C}>\angle\text{B},$ then:
→- BC > AC
- AB > AC
- AB < AC
- BC < AC
The graph of y = 4x will: