MCQ
- A$45^{\circ}$
- B$55^{\circ}$
- ✓$60^{\circ}$
- D$75^{\circ}$
✓
Answer
Correct option: C.
$60^{\circ}$
(c) $60^{\circ}$
We have $\angle A=75^{\circ}$. Therefore, $\angle C=75^{\circ}$.
Using angle sum property in $\triangle B C D$, we obtain: $\angle C B D=60^{\circ}$.
We have $\angle A=75^{\circ}$. Therefore, $\angle C=75^{\circ}$.
Using angle sum property in $\triangle B C D$, we obtain: $\angle C B D=60^{\circ}$.
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
After rationalising the denominator of $\frac{7}{3\sqrt{3}-2\sqrt{2}},$ we get the denominator as.
→The mean of n observations is $\overline{\text{X}}.$ If $k$ is added to each observation, then the new mean is:
→If $\text{x}=3+2\sqrt{2},$ than the value of $\text{x}+\frac{1}{\text{x}}$ is:
→If $x = 3$ and $y = -2$ satisfies $2x - 3y = k$, then the value of $k$ is:
→If $(x+y)^3-(x-y)^3-6 y\left(x^2-y^2\right)=k y^2$, then $k=$
→The rationalisation factor of $\sqrt3,$ is:
→Two chords $AB$ and $CD$ of a circle intersect each other at a point $E$ outside the circle. If $AB = 11\ cm, BE = 3\ cm$ and $DE = 3.5\ cm,$ then $CD = ?$
→If $ABC$ is an arc of a circle and $\angle\text{ABC} = 135^\circ,$ then the ratio of arc $\widehat{\text{ABC}}$ to the circumference is:
→The value of $\frac{\left(a^2-b^2\right)^3+\left(b^2-c^2\right)^3+\left(c^2-a^2\right)^3}{(a-b)^3+(b-c)^3+(c-a)^3}$ is
→The total surface area of a cone whose radius is $\frac{\text{r}}{2}$ and slant height $2l$ is:
→