Segments on the hypotenuse

A right triangle ABC has a hypotenuse c = 26 cm. The altitude from C to the hypotenuse is h_c = 12 cm. What are the lengths of the two segments of the hypotenuse? What are the lengths of sides a and b? What are the angles at vertices A and B?

Final Answer:

c₁ =  18 cm
c₂ =  8 cm
a =  21.6333 cm
b =  14.4222 cm
α =  56.3099 °
β =  33.6901 °

Step-by-step explanation:

$$\begin{gathered} c=26\text{cm} \\ v=12\text{cm} \\ c=c_1+c_2 \\ {c}_1\cdot c_2=v^2 \\ {c}_1\cdot (c-c_1)=v^2 \\ x\cdot (c-x)=v^2 \\ x\cdot (26-x)=12^2 \\ -x^2+26x-144=0 \\ {x}^2-26x+144=0 \\ a=1;b=-26;c=144 \\ D=b^2-4ac=26^2-4\cdot 1\cdot 144=100 \\ D>0 \\ {x}_{1,2}=\frac{-b\pm \sqrt{D}}{2a}=\frac{26\pm \sqrt{100}}{2} \\ {x}_{1,2}=\frac{26\pm 10}{2} \\ {x}_{1,2}=13\pm 5 \\ {x}_1=18 \\ {x}_2=8 \\ {c}_1=x_1=18=18\text{cm} \end{gathered}$$

Our quadratic equation calculator calculates it.

$c_2=c-c_1=26-18=8\text{cm}$

$$\begin{gathered} a^2=c_1\cdot c \\ a=\sqrt{c_1\cdot c}=\sqrt{18\cdot 26}=6\sqrt{13}=21.6333\text{cm} \end{gathered}$$

$$\begin{gathered} b^2=c_2\cdot c \\ b=\sqrt{c_2\cdot c}=\sqrt{8\cdot 26}=4\sqrt{13}=14.4222\text{cm} \end{gathered}$$

$$\begin{gathered} \sin \alpha =a:c=21.6333:26\doteq 0.8321=134996:162245 \\ \alpha =\frac{180°}{\pi }\cdot \arcsin (a/c)=\frac{180°}{\pi }\cdot \arcsin (21.6333/26)=56.3099^{\circ }=56°18^{\prime }36" \end{gathered}$$

$$\begin{gathered} \sin \beta =b:c=14.4222:26\doteq 0.5547=92159:166142 \\ \beta =\frac{180°}{\pi }\cdot \arcsin (b/c)=\frac{180°}{\pi }\cdot \arcsin (14.4222/26)\doteq 33.6901^{\circ } \\ \text{ Verifying Solution: } \\ C=\sqrt{a^2+b^2}=\sqrt{21.6333^2+14.4222^2}=26\text{cm} \\ C=c \end{gathered}$$

Try another example