Question

I need help urgently
Solve the whole right triangle

Answer 1

Angles:

• m∠A = 51.3°
• m∠B = 38.7°
• m∠C = 90°

Sides:

• AB = 16 units
• BC = 2√(39) units ≈ 12.5 units
• AC = 10 units

Explanation:

When "solving a triangle" we need to find the measures of all three interior angles, and the lengths of all three sides of the triangle.

Angles

Since the triangle is a right triangle, we can use trigonometric ratios to find the measure of one of the unknown angles. Side AC is adjacent to angle A and side AB is the hypotenuse of the triangle, so we can use the cosine trigonometric ratio.

`\boxed{\begin{array}{l}\underline{\textsf{Cosine trigonometric ratio}}\\\\\sf \cos(\theta)=(A)/(H)\\\\\textsf{where:}\\\phantom{ww}\bullet\;\textsf{$\theta$ is the angle.}\\\phantom{ww}\bullet\;\textsf{$A$ is the side adjacent the angle.}\\\phantom{ww}\bullet\;\textsf{$H$ is the hypotenuse (the side opposite the right angle).}\end{array}}`

Given than AC = 10 and AB = 16:

`\begin{aligned}\cos A&=(AC)/(AB)\\\\\cos A&=(10)/(16)\\\\\cos A&=(5)/(8)\\\\A&=\cos^(-1)\left((5)/(8)\right)\\\\A&=51.31781254651...^(\circ)\\\\A&=51.3^(\circ)\; \sf (nearest\;tenth)\end{aligned}`

Now, use the fact that the interior angles of a triangle sum to 180° to determine the measure of angle B.

`\begin{aligned}m \angle A+m \angle B+m \angle C&=180^(\circ)\\51.3^(\circ)+m \angle B+90^(\circ)&=180^(\circ)\\m \angle B+141.3^(\circ)&=180^(\circ)\\m \angle B&=38.7^(\circ)\end{aligned}`

Therefore, the measures of all three angles of the right triangle are:

• m∠A = 51.3°
• m∠B = 38.7°
• m∠C = 90°

Side lengths

As the triangle is a right triangle, we can use Pythagoras Theorem to find the measure of leg BC (the leg opposite angle A).

`\boxed{\begin{array}{l}\underline{\sf Pythagoras \;Theorem} \\\\\Large\text{$a^2+b^2=c^2$}\\\\\textsf{where:}\\\phantom{ww}\bullet\;\textsf{$a$ and $b$ are the legs of the right triangle.}\\\phantom{ww}\bullet\;\textsf{$c$ is the hypotenuse (longest side) of the right triangle.}\\\end{array}}`

Given AC = 10 and AB = 16, substitute these values into the formula and solve for BC:

`\begin{aligned}BC^2+AC^2&=AB^2\\BC^2&=AB^2-AC^2\\BC&=√(AB^2-AC^2)\\BC&=√(16^2-10^2)\\BC&=√(256-100)\\BC&=√(156)\\BC&=√(2^2\cdot39)\\BC&=√(2^2)√(39)\\BC&=2√(39)\end{aligned}`

Therefore, the lengths of all three sides of the right triangle are:

• AB = 16 units
• BC = 2√(39) units ≈ 12.5 units
• AC = 10 units