Question
Find the measure of an angle which is 36º more than its complement.
✓
Answer
Let the measure of the required angle be xº.
then, measure of its complement = (90 - x)º.
Therefore,
x - (90º - x) = 36º
⇒ 2x = 126º
⇒ x = 63º
Hence, the measure of the required angle is 63º.
then, measure of its complement = (90 - x)º.
Therefore,
x - (90º - x) = 36º
⇒ 2x = 126º
⇒ x = 63º
Hence, the measure of the required angle is 63º.
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
Follow the arrows and complete the chart by doing the operations given.
Solve for $\text{x}\Big(\frac{2}{5}\Big)^{2\text{x}-2}=\frac{32}{3125}.$
→Evaluate:$\frac{2^0+7^0}{5^0}$
→In the adjoining figure, show that$\angle\text{A}+\angle\text{B}+\angle\text{C}+\angle\text{D}+\angle\text{E}+\angle\text{F}=360^\circ.$
→In figure, lines AB, CD and EF intersect at O. Find the measures of $\angle\text{AOC},\angle\text{COF},\angle\text{DOE}$ and $\angle\text{BOF}.$
→Write the following in decimal form and say what kind of decimal expansion has.$\frac{15}{13}$
→Simplify the following:$\frac{(\text{a}^{3\text{n-9}})^6}{\text{a}^{2\text{n}-4}}$
→Solve the following equations: $4^{\text{x}-1}\times(0.5)^{3-2\text{x}}=\Big(\frac{1}{8}\Big)^{\text{x}}$
→Tan θ =$\frac{21}{20}$.Find Sin θ= , Cos θ= ?
→Factorise:
$27 a^3-b^3+8 c^3-18 a b c$
→$27 a^3-b^3+8 c^3-18 a b c$