Question

Solving a Right

Triangle Solve the right triangle

Short answer

Base of triangle is $\frac{\mathbf{100}\sqrt{\mathbf{2}}}{\mathbf{2}\mathbf{-}\sqrt{\mathbf{3}}}$, the hypotenuse is $\mathbf{100}\mathbf{(}\sqrt{\mathbf{6}}\mathbf{-}\sqrt{\mathbf{2}}\mathbf{)}$ and the remaining angle is${\mathbf{15}}^{\mathbf{\circ }}$.

Step-by-step solution

Step 1. Given information

A right triangle with known value of a side of 100 and an angle of${75}^{\circ }$

Step 2. Concept used

A side and an angle are both known. Use trigonometric formulas to get the unknown side. And use triangle sum property of angles to find unknown angle. Use formulae as-

$\begin{array}{l}\Rightarrow \sin \theta =\frac{p}{h}\\ \Rightarrow \cos \theta =\frac{b}{h}\\ \Rightarrow \tan \theta =\frac{p}{b}\end{array}$

Where

$\begin{array}{l}h=hypotenuse\\ b=base\\ p=perpendicular\end{array}$

Step 3. Calculation

Now use the formulae,

$\begin{array}{l}\sin \theta =\frac{p}{h}\\ \because \sin {75}^{°}=\frac{\sqrt{6}+\sqrt{2}}{4}\\ \Rightarrow \frac{\sqrt{6}+\sqrt{2}}{4}=\frac{100}{\text{h}}\\ \Rightarrow h=\frac{400}{\sqrt{6}+\sqrt{2}}\\ =\frac{400(\sqrt{6}-\sqrt{2})}{(\sqrt{6}+\sqrt{2})(\sqrt{6}-\sqrt{2})}\\ =\frac{400(\sqrt{6}-\sqrt{2})}{6-2}\\ =\frac{400(\sqrt{6}-\sqrt{2})}{4}\\ =100(\sqrt{6}-\sqrt{2})\end{array}$

Use sum of triangle’s angle,

$\begin{array}{l}\angle A+\angle B+\angle C={180}^{\circ }\\ \angle A+75+90=180\\ \angle A=180-165\\ \angle A={15}^{\circ }\end{array}$

The remaining side,

$\begin{array}{l}\tan \theta =\frac{p}{b}\\ \tan 15=\frac{100}{b}\\ \frac{2-\sqrt{3}}{\sqrt{2}}=\frac{100}{b}\\ b=\frac{100\sqrt{2}}{2-\sqrt{3}}\end{array}$