Question

please help me this question ​

Answer 1

${\huge{\underline{\boxed{\red{\mathcal{Answer}}}}}}$

To Evaluate:

• $\dfrac{5\cos^260^{\circ} + 4\sec^230^{\circ} - \tan^245^{\circ}}{\sin^245^{\circ} + \cos^245^{\circ}}$

Solution:

$\dfrac{5\cos^260^{\circ} + 4\sec^230^{\circ} - \tan^245^{\circ}}{\sin^245^{\circ} + \cos^245^{\circ}} \\\\\text{Substituting the values of the given trigonometric ratios, we get,}$

$\implies \dfrac{5\left(\dfrac{1}{2}\right)^2 + 4\left(\dfrac{2}{\sqrt{3}}\right)^2 - (1)^2}{\left(\dfrac{1}{\sqrt{2}}\right)^2 + \left(\dfrac{1}{\sqrt{2}}\right)^2}$

$\implies \dfrac{5*\dfrac{1}{4} + 4*\dfrac{4}{3} - 1}{\dfrac{1}{2} + \dfrac{1}{2}}$

$\implies \dfrac{\dfrac{5}{4} + \dfrac{16}{3} - 1}{1} \\\\\implies \dfrac{(5*3) + (4*16) - (12*1)}{12} \:\:\:\:\:Take\:LCM\\\\\implies \dfrac{15 + 64 - 12}{12} \\\\\bf{\implies \dfrac{67}{12}}$

Learn More:

$\large{\sf Trigonometric Ratios-:} \\ \Large{ \begin{tabular}{|c|c|c|c|c|c|} \cline{1-6} \theta & \sf 0^{\circ} & \sf 30^{\circ} & \sf 45^{\circ} & \sf 60^{\circ} & \sf 90^{\circ} \\ \cline{1-6} \sin & 0 & \dfrac{1}{2 } & \dfrac{1}{ \sqrt{2} } & \dfrac{ \sqrt{3}}{2} & 1 \\ \cline{1-6} \cos & 1 & \dfrac{ \sqrt{ 3 }}{2} } & \dfrac{1}{ \sqrt{2} } & \dfrac{ 1 }{ 2 } & 0 \\ \cline{1-6} \tan & 0 & \dfrac{1}{ \sqrt{3} } & 1 & \sqrt{3} & \infty \\ \cline{1-6} \cot & \infty & \sqrt{3} & 1 & \dfrac{1}{ \sqrt{3} } &0 \\ \cline{1 - 6} \sec & 1 & \dfrac{2}{ \sqrt{3}} & \sqrt{2} & 2 & \infty \\ \cline{1-6} \csc & \infty & 2 & \sqrt{2 } & \dfrac{ 2 }{ \sqrt{ 3 } } & 1 \\ \cline{1 - 6}\end{tabular}}$

Hope it helps!!!

Answer 2

Answer:

From the question it is given that,

EF = 7.2 cm

∠E = 110o

∠F = 80o

Now we have to check whether it is possible to construct ΔDEF from the given values.

We know that the sum of the angles of a triangle is 180o.

Then,

∠D + ∠E + ∠F = 180o

∠D + 110o+ 80o= 180o

∠D + 190o = 180o

∠D = 180o– 1900

∠D = -10o

We may observe that the sum of two angles is 190o is greater than 180o. So, it is not possible to construct a triangle.