# Question Video: Finding the Sine and Tangent of Angles in Right Triangles Where the Hypotenuse Is Twice the Opposite Side Mathematics

Find sin π΄ and tan π΄, given that π΄π΅πΆ is a right triangle at π΅, where 2πΆπ΅ = π΄πΆ.

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### Video Transcript

Find sin π΄ and tan π΄, given that π΄π΅πΆ is a right triangle at π΅, where two πΆπ΅ equals π΄πΆ.

Weβre given information about this right triangle π΄π΅πΆ, so letβs begin by sketching it out. The right triangle is at π΅, so, in other words, sides π΄π΅ and π΅πΆ are perpendicular. Weβre also told that two πΆπ΅ is equal to π΄πΆ, so letβs let the side πΆπ΅ be equal to π₯ length units. Then, side π΄πΆ must be equal to two π₯ length units. Now, weβre going to be calculating sin of π΄ and tan of π΄. And we know for a right triangle with included angle π, the trigonometric ratios sine and tangent tell us that sin of π is equal to the length of the opposite divided by the length of the hypotenuse. And tan of π is equal to opposite divided by adjacent.

Now, weβre trying to find sin of π΄ and tan of π΄, so thatβs the sine and tangent of this angle here. Relative to this included angle, we know the length of the opposite. Itβs the length of π΅πΆ; itβs π₯ units. And the hypotenuse in this triangle is two π₯ units. This means we can quite quickly calculate the value of sin of π΄. But we do need to do a little bit more work to find the value of tan of π΄.

Letβs begin by finding the value of sin of π΄. Itβs the length of the opposite divided by the length of the hypotenuse. Thatβs π₯ divided by two π₯. But of course, we can simplify this expression by dividing through by π₯. So π₯ divided by two π₯ simplifies to one-half, and sin π΄ is equal to one-half. tan of π΄, of course, is the opposite divided by the adjacent. So letβs find an expression for the length of the adjacent side.

To do so, weβll use the Pythagorean theorem. And this tells us that the sum of the squares of the two shorter sides in our triangle must be equal to the square of the hypotenuse. In other words, π₯ squared plus π΄π΅ squared must be equal to two π₯ squared. Two π₯ all squared is equal to four π₯ squared. So weβll now make π΄π΅ squared the subject by subtracting π₯ squared from both sides. π΄π΅ squared is equal to three π₯ squared then, meaning that π΄π΅ is equal to the positive square root of three π₯ squared. This can alternatively be written as the square root of three times π₯. So the length of π΄π΅ is root three π₯ units.

Weβre now able to find an expression for tan of π΄. Since itβs opposite over adjacent, itβs π₯ over root three π₯. We can once again simplify by dividing through by π₯, giving us one over root three. Then, we can simplify further by rationalizing the denominator of this fraction. To do so, we multiply both the numerator and denominator by the square root of three. Then, root three times root three is three. So we find that tan of π΄ is equal to root three over three. And we found the values of sin of π΄ and tan of π΄. sin π΄ is one-half and tan π΄ is root three over three.