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Right Triangle Word Problems

A triangle is a polygon of three sides. In fact it is the first member in the family of polygons. Triangles are classified as different types based on measures of sides, measures of interior angles etc. A right triangle is one of the special types of a triangle.

 Related Calculators Area of a Right Triangle Calculator

Definition

A right triangle is a triangle in which measure of one interior angle is $90$ degrees. That angle is also called as right angle and thus it is named as right angle triangle or just as right triangle. The sides that make the right angle are called 'legs' and the remaining side which is the longest and opposite to the right angle is called as 'hypotenuse'. Only in a right triangle a relation can be established between the measures of all the sides without the knowledge of the remaining interior angles. That is, if '$a$', '$b$' and '$c$' are the measures of the legs of the right triangle and if '$c$' is the measure of hypotenuse, then, $c^2$ = $a^2 + b^2$. This is true for any right triangle irrespective of the magnitudes of the sides. This was discovered by the very famous Greek mathematician Pythagoras as back as about $2500$ years ago. This statement is described as Pythagorean Theorem after his name and it is applied in many real life situations.

A right triangle is the origin for the topic of trigonometry. In a right triangle, the ratios of different pairs of sides are constants for a given base angle and such ratios are called trigonometric ratios for that angle. The ratio of the corresponding opposite side to angle '$x$' and the hypotenuse is defined as Sine ratio of angle '$x$' and it is denoted as sin $(x)$. The ratio of adjacent side and hypotenuse is Cosine ratio, denoted as $cos (x)$ and the ratio of opposite side and adjacent side is Tangent ratio and denoted as $tan(x)$.

These ratio values for all angles can be found in mathematical tables and in scientific calculators. Such values are used in solving different problems of practical situations, especially related to heights and distances.

Word Problems

Example 1:

A ladder is $13$ feet long and its foot is placed $5$ feet away from the bottom of a vertical wall. Up to what height on the wall, the top of the ladder reaches?

Solution:

A simple sketch will help to easily solve the problem. Assuming the floor is horizontal, it forms a right angle with the wall. Let $A$ be the top of the ladder $B$ be its bottom. The angle $BCA$ is right angle. Applying Pythagorean Theorem,

$BC^2 + CA^2$ = $AB^2$,  or  $5^2 + h^2$ = $132$,  or  $25 + h^2$ = $169$,  or   $h^2$ = $144$ or, $h$ = $12$

So the top of the ladder will be at an height of $12 ft$.
Example 2:

A rectangular garden has a length which is $5 ft$ more than twice the width. If the diagonal of the garden measures $85$ feet, find the length and width of the garden.

Solution:

One set of sides as legs and the diagonal of the rectangular garden as hypotenuse form a right triangle. Let '$x$' feet be the width of the garden. Then, as per the problem statement, the length is $(2x + 5)$ feet. The diagonal is given as $85$ feet.

Now, applying Pythagorean Theorem,

$x^2 + (2x + 5)^2$ = $85^2$ or $x^2 + 4x^2 + 20x + 25$ = $7225$ or $5x^2 + 20x +25$ = $7225$ or $5x^2 + 20x - 7200$ = $0$

Now, dividing throughout by $5$, the equation becomes as,

$x^2 + 4x - 1440$ = $0$ or $(x + 40)(x â€“ 36)$ = $0$. It gives the solutions as $x$ = $-40$ and $x$ = $36$. Since a negative solution cannot be an answer in this situation, $x$ = $36$ is the only solution. In such a case, the width of the garden is $36$ feet and the length of the garden is $(2)(36) + 5$ = $77$ feet.
Example 3:

A lamp post is held by two stay wires; both are tied at the same point on the post. The other ends are pegged on the ground. The length of shorter wire is $25$ feet and that of the longer one is $26$ feet. If the distance between the pegs on the ground is $3$ feet, find at what height on the lamp post both the wires are tied.

Look at the following diagram. Solution:

Let $A$ be the point where the wires are tied at the top of the lamp post and $B$ be its bottom at the ground. The first nearer peg is at $C$ and the farther is at $D$. As per the problem statement, $AC$ = $25$ feet and $AD$ = $26$ feet. Let $AB$ = $h$ feet and $BC$ = $x$ feet. Then $BD$ would be $(x + 3)$ feet.

Clearly both the triangle $ABC$ and $ABD$ are right triangles with right angle at $B$. Therefore, Pythagorean Theorem can be applied on both these triangles.

In triangle $ABD,\ h^2\ +\ (x + 3)^2$ = $26^2$ â€¦ (1)

In triangle $ABC,\ h^2\ +\ x^2$ = $25^2$ â€¦ (2)

Subtracting (2) from (1), $(x + 3)^2\ - x^2$ = $26^2 - 25^2$

Using the difference of squares identity, $(x + 3 + x)(x + 3 - x)$ = $(26 + 25)(26 - 25)$

Or, $(2x + 3)(3)$ = $51$ or, $2x + 3$ = $17$ or $2x$ = $14$ or $x$ = $7$

Plugging in $x$ = $7$ in equation $(2),\ h^2 + 7^2$ = $25^2$ or $h^2 + 49$ = $625$  or $h^2$ = $576$ or $h$ = $24$

So, the both the wires are tied at an height of $24$ feet on the lamp post.
Example 4:

When the Sun is at an angle of $60^{\circ}$ to the horizon, the shadow of a building is noted as $30$ feet on a horizontal ground. Find the height of the building to the nearest foot.

Draw a sketch of the situation. Solution:

The Sun ray, the tree and its shadow form a right triangle with the base angle at the ground as $60^{\circ}$. Let '$h$' be the height of the tree and '$s$' be the length of shadow.

Applying the concept trigonometric ratios,

$tan\ (60^{\circ})$ = $\frac{h}{s}$ or $h$ = $s \times tan\ (60^{\circ})$ = $30 \times\ tan\ (60^{\circ})$

Using a scientific calculator we can find $h \approx 52$ feet.
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